前言
前两天小编刚忙完手头上的事情，闲了下来，然后顺便研究了一下Branch and Price的算法。刚好，国内目前缺少这种类型算法的介绍和代码实现，今天就给大家分享一下咯。
代码出自国外一个大神@author mschyns之手。代码没有写调用模块，这一部分是小编后续补上去的，以便大家能运行（真是太贴心啦呜呜呜~还不赶紧转发点赞！）。然后检查了代码，修正了一些bug。
代码获取方式：关注公众号！
在公众号后台回复【BPVRPTW】不包括【】即可。
算法介绍
该Branch and Price算法由以下几部分组成：
-
Branch and Bound：分支定界，下界使用Column Generation求解。
-
Column Generation：列生成算法，求解VRPWTW松弛模型的最优解。
-
ESPPRC-Label Setting：求解VRPTW的子问题(pricing problem)，标号法求解。
算法的运行效果如下：
算例用的是标准Solomon25。大部分，一轮Column Generation就能直接得到整数解，可能是巧合。也有部分算例需要branch。
更改输入算例在Main.java:
更改算例后同时也要更改客户数，在paramsVRP.java:
可参考的推文如下
CPLEX：
1. 干货 | cplex介绍、下载和安装以及java环境配置和API简单说明
2. 干货 | JAVA调用cplex求解一个TSP模型详解
3. 干货|十分钟快速掌握CPLEX求解VRPTW数学模型（附JAVA代码及CPLEX安装流程）
Branch and Bound
1. 干货 | 10分钟带你全面掌握branch and bound（分支定界）算法-概念篇
2. 干货 | 10分钟搞懂branch and bound算法的代码实现附带java代码
3. 干货 | 10分钟教你用branch and bound（分支定界）算法求解TSP旅行商问题
4. cplex教学 | 分支定界法(branch and bound)解带时间窗的车辆路径规划问题（附代码及详细注释）
Column Generation
1. 干货 | 10分钟带你彻底了解Column Generation（列生成）算法的原理附java代码
2. 运筹学教学|列生成（Column Generation）算法（附代码及详细注释）
3. 干货 | 10分钟教你使用Column Generation求解VRPTW的线性松弛模型
4. 干货 | 求解VRPTW松弛模型的Column Generation算法的JAVA代码分享
ESPPRC
1. 干货 | VRPTW子问题ESPPRC的介绍及其求解算法的C++代码
2. 标号法(label-setting algorithm)求解带时间窗的最短路问题
可参考的文献如下
BOOK: Desrosiers, Desaulniers, Solomon, "Column Generation", Springer, 2005 (GERAD, 25th anniversary)
Column generation : chapter 3, Vehicle Routing Problem with Time Windows Brian Kallehauge, Jesper Larsen, Oli B. G, Madsen, and Marius M. Solomon
Pricing SPPRC : chapter 2, Shortest Path Problems with Resource Constraints 33 Stefan Irnich and Guy Desaulniers
文献下载请关注公众号！
在后台回复【BPREF】不包括【】即可下载。
代码解析如下
Branch and Bound的过程如下（具体参考此前讲过的算法原理）：
public boolean BBnode(paramsVRP userParam, ArrayList<route> routes,
treeBB branching, ArrayList<route> bestRoutes, int depth)
throws IOException {
// userParam (input) : all the parameters provided by the users (cities,
// roads...)
// routes (input) : all (but we could decide to keep only a subset) the
// routes considered up to now (to initialize the Column generation process)
// branching (input): BB branching context information for the current node
// to process (branching edge var, branching value, branching from...)
// bestRoutes (output): best solution encountered
int i, j, bestEdge1, bestEdge2, prevcity, city, bestVal;
double coef, bestObj, change, CGobj;
boolean feasible;
try {
// check first that we need to solve this node. Not the case if we have
// already found a solution within the gap precision
if ((upperbound - lowerbound) / upperbound < userParam.gap)
return true;
// init
if (branching == null) { // root node - first call
// first call - root node
treeBB newnode = new treeBB();
newnode.father = null;
newnode.toplevel = true;
newnode.branchFrom = -1;
newnode.branchTo = -1;
newnode.branchValue = -1;
newnode.son0 = null;
branching = newnode;
}
// display some local info
if (branching.branchValue < 1)
System.out.println("
Edge from " + branching.branchFrom + " to "
+ branching.branchTo + ": forbid");
else
System.out.println("
Edge from " + branching.branchFrom + " to "
+ branching.branchTo + ": set");
int mb = 1024 * 1024;
Runtime runtime = Runtime.getRuntime();
System.out.print("Java Memory=> Total:" + (runtime.totalMemory() / mb)
+ " Max:" + (runtime.maxMemory() / mb) + " Used:"
+ ((runtime.totalMemory() - runtime.freeMemory()) / mb) + " Free: "
+ runtime.freeMemory() / mb);
// Compute a solution for this node using Column generation
columngen CG = new columngen();
CGobj = CG.computeColGen(userParam, routes);
// feasible ? Does a solution exist?
if ((CGobj > 2 * userParam.maxlength) || (CGobj < -1e-6)) {
// can only be true when the routes in the solution include forbidden edges (can happen when the BB set branching values)
System.out.println("RELAX INFEASIBLE | Lower bound: " + lowerbound
+ " | Upper bound: " + upperbound + " | Gap: "
+ ((upperbound - lowerbound) / upperbound) + " | BB Depth: "
+ depth + " | " + routes.size() + " routes");
return true; // stop this branch
}
branching.lowestValue = CGobj;
// update the global lowerbound when required
if ((branching.father != null) && (branching.father.son0 != null)
&& branching.father.toplevel) {
// all nodes above and on the left have been processed=> we can compute
// a new lowerbound
lowerbound = (branching.lowestValue > branching.father.son0.lowestValue) ? branching.father.son0.lowestValue
: branching.lowestValue;
branching.toplevel = true;
} else if (branching.father == null) // root node
lowerbound = CGobj;
if (branching.lowestValue > upperbound) {
CG = null;
System.out.println("CUT | Lower bound: " + lowerbound
+ " | Upper bound: " + upperbound + " | Gap: "
+ ((upperbound - lowerbound) / upperbound) + " | BB Depth: "
+ depth + " | Local CG cost: " + CGobj + " | " + routes.size()
+ " routes");
return true; // cut this useless branch
} else {
// ///////////////////////////////////////////////////////////////////////////
// check the (integer) feasibility. Otherwise search for a branching
// variable
feasible = true;
bestEdge1 = -1;
bestEdge2 = -1;
bestObj = -1.0;
bestVal = 0;
// transform the path variable (of the CG model) into edges variables
for (i = 0; i < userParam.nbclients + 2; i++)
java.util.Arrays.fill(userParam.edges[i], 0.0);
for (route r : routes) {
if (r.getQ() > 1e-6) { // we consider only the routes in the current
// local solution
ArrayList<Integer> path = r.getpath(); // get back the sequence of
// cities (path for this
// route)
prevcity = 0;
for (i = 1; i < path.size(); i++) {
city = path.get(i);
userParam.edges[prevcity][city] += r.getQ(); // convert into edges
prevcity = city;
}
}
}
// find a fractional edge
for (i = 0; i < userParam.nbclients + 2; i++) {
for (j = 0; j < userParam.nbclients + 2; j++) {
coef = userParam.edges[i][j];
if ((coef > 1e-6)
&& ((coef < 0.9999999999) || (coef > 1.0000000001))) {
// this route has a fractional coefficient in the solution =>
// should we branch on this one?
feasible = false;
// what if we impose this route in the solution? Q=1
// keep the ref of the edge which should lead to the largest
// change
change = (coef < Math.abs(1.0 - coef)) ? coef : Math.abs(1.0 - coef);
change *= routes.get(i).getcost();
if (change > bestObj) {
bestEdge1 = i;
bestEdge2 = j;
bestObj = change;
bestVal = (Math.abs(1.0 - coef) > coef) ? 0 : 1;
}
}
}
}
CG = null;
if (feasible) {
if (branching.lowestValue < upperbound) { // new incumbant feasible solution!
upperbound = branching.lowestValue;
bestRoutes.clear();
for (route r : routes) {
if (r.getQ() > 1e-6) {
route optim = new route();
optim.setcost(r.getcost());
optim.path = r.getpath();
optim.setQ(r.getQ());
bestRoutes.add(optim);
}
}
System.out.println("OPT | Lower bound: " + lowerbound
+ " | Upper bound: " + upperbound + " | Gap: "
+ ((upperbound - lowerbound) / upperbound) + " | BB Depth: "
+ depth + " | Local CG cost: " + CGobj + " | " + routes.size()
+ " routes");
System.out.flush();
} else
System.out.println("FEAS | Lower bound: " + lowerbound
+ " | Upper bound: " + upperbound + " | Gap: "
+ ((upperbound - lowerbound) / upperbound) + " | BB Depth: "
+ depth + " | Local CG cost: " + CGobj + " | " + routes.size()
+ " routes");
return feasible;
} else {
System.out.println("INTEG INFEAS | Lower bound: " + lowerbound
+ " | Upper bound: " + upperbound + " | Gap: "
+ ((upperbound - lowerbound) / upperbound) + " | BB Depth: "
+ depth + " | Local CG cost: " + CGobj + " | " + routes.size()
+ " routes");
System.out.flush();
// ///////////////////////////////////////////////////////////
// branching (diving strategy)
// first branch -> set edges[bestEdge1][bestEdge2]=0
// record the branching information in a tree list
treeBB newnode1 = new treeBB();
newnode1.father = branching;
newnode1.branchFrom = bestEdge1;
newnode1.branchTo = bestEdge2;
newnode1.branchValue = bestVal; // first version was not with bestVal
// but with 0
newnode1.lowestValue = -1E10;
newnode1.son0 = null;
// branching on edges[bestEdge1][bestEdge2]=0
EdgesBasedOnBranching(userParam, newnode1, false);
// the initial lp for the CG contains all the routes of the previous
// solution less the routes containing this arc
ArrayList<route> nodeRoutes = new ArrayList<route>();
for (route r : routes) {
ArrayList<Integer> path = r.getpath();
boolean accept = true;
if (path.size() > 3) { // we must keep trivial routes
// Depot-City-Depot in the set to ensure
// feasibility of the CG
prevcity = 0;
for (j = 1; accept && (j < path.size()); j++) {
city = path.get(j);
if ((prevcity == bestEdge1) && (city == bestEdge2))
accept = false;
prevcity = city;
}
}
if (accept)
nodeRoutes.add(r);
}
boolean ok;
ok = BBnode(userParam, nodeRoutes, newnode1, bestRoutes, depth + 1);
nodeRoutes = null; // free memory
if (!ok) {
return false;
}
branching.son0 = newnode1;
// second branch -> set edges[bestEdge1][bestEdge2]=1
// record the branching information in a tree list
treeBB newnode2 = new treeBB();
newnode2.father = branching;
newnode2.branchFrom = bestEdge1;
newnode2.branchTo = bestEdge2;
newnode2.branchValue = 1 - bestVal; // first version: always 1
newnode2.lowestValue = -1E10;
newnode2.son0 = null;
// branching on edges[bestEdge1][bestEdge2]=1
// second branching=>need to reinitialize the dist matrix
for (i = 0; i < userParam.nbclients + 2; i++)
System.arraycopy(userParam.distBase[i], 0, userParam.dist[i], 0,
userParam.nbclients + 2);
EdgesBasedOnBranching(userParam, newnode2, true);
// the initial lp for the CG contains all the routes of the previous
// solution less the routes incompatible with this arc
ArrayList<route> nodeRoutes2 = new ArrayList<route>();
for (route r : routes) {
ArrayList<Integer> path = r.getpath();
boolean accept = true;
if (path.size() > 3) { // we must keep trivial routes
// Depot-City-Depot in the set to ensure
// feasibility of the CG
prevcity = 0;
for (i = 1; accept && (i < path.size()); i++) {
city = path.get(i);
if (userParam.dist[prevcity][city] >= userParam.verybig - 1E-6)
accept = false;
prevcity = city;
}
}
if (accept)
nodeRoutes2.add(r);
}
ok = BBnode(userParam, nodeRoutes2, newnode2, bestRoutes, depth + 1);
nodeRoutes2 = null;
// update lowest feasible value of this node
branching.lowestValue = (newnode1.lowestValue < newnode2.lowestValue) ? newnode1.lowestValue
: newnode2.lowestValue;
return ok;
}
}
} catch (IOException e) {
System.err.println("Error: " + e);
}
return false;
}
Column Generation的过程如下，Master Problem采用vrptw的set covering model 的松弛模型，利用cplex建模求解，求解的结果作为branch and bound的lower bound：
public double computeColGen(paramsVRP userParam, ArrayList<route> routes)
throws IOException {
int i, j, prevcity, city;
double cost, obj;
double[] pi;
boolean oncemore;
try {
// ---------------------------------------------------------
// construct the model for the Restricted Master Problem
// ---------------------------------------------------------
// warning: for clarity, we create a new cplex env each time we start a
// Column Generation
// this class contains (nearly) everything about CG and could be used
// independently
// However, since the final goal is to encompass it inside 锟� Branch and
// Bound (BB),
// it would (probably) be better to create only once the CPlex env when we
// initiate the BB and to work with the same (but adjusted) lp matrix each
// time
IloCplex cplex = new IloCplex();
IloObjective objfunc = cplex.addMinimize();
// for each vertex/client, one constraint (chapter 3, 3.23 )
IloRange[] lpmatrix = new IloRange[userParam.nbclients];
for (i = 0; i < userParam.nbclients; i++)
lpmatrix[i] = cplex.addRange(1.0, Double.MAX_VALUE);
// for each constraint, right member >=1
// lpmatrix[i] = cplex.addRange(1.0, 1.0);
// or for each constraint, right member=1 ... what is the best?
// Declaration of the variables
IloNumVarArray y = new IloNumVarArray(); // y_p to define whether a path p
// is used
// Populate the lp matrix and the objective function
// first with the routes provided by the argument 'routes' of the function
// (in the context of the Branch and Bound, it would be a pity to start
// again the CG from scratch at each node of the BB!)
// (we should reuse parts of the previous solution(s))
for (route r : routes) {
int v;
cost = 0.0;
prevcity = 0;
for (i = 1; i < r.getpath().size(); i++) {
city = r.getpath().get(i);
cost += userParam.dist[prevcity][city];
prevcity = city;
}
r.setcost(cost);
IloColumn column = cplex.column(objfunc, r.getcost());
// obj coefficient
for (i = 1; i < r.getpath().size() - 1; i++) {
v = r.getpath().get(i) - 1;
column = column.and(cplex.column(lpmatrix[v], 1.0));
// coefficient of y_i in (3.23) => 0 for the other y_p
}
y.add(cplex.numVar(column, 0.0, Double.MAX_VALUE));
// creation of the variable y_i
}
// complete the lp with basic route to ensure feasibility
if (routes.size() < userParam.nbclients) { // a priori true only the first time
for (i = 0; i < userParam.nbclients; i++) {
cost = userParam.dist[0][i + 1]
+ userParam.dist[i + 1][userParam.nbclients + 1];
IloColumn column = cplex.column(objfunc, cost); // obj coefficient
column = column.and(cplex.column(lpmatrix[i], 1.0)); // coefficient of
// y_i in (3.23)
// => 0 for the
// other y_p
y.add(cplex.numVar(column, 0.0, Double.MAX_VALUE)); // creation of the
// variable y_i
route newroute = new route();
newroute.addcity(0);
newroute.addcity(i + 1);
newroute.addcity(userParam.nbclients + 1);
newroute.setcost(cost);
routes.add(newroute);
}
}
// cplex.exportModel("model.lp");
// CPlex params
cplex.setParam(IloCplex.IntParam.RootAlg, IloCplex.Algorithm.Primal);
cplex.setOut(null);
// cplex.setParam(IloCplex.DoubleParam.TiLim,30); // max number of
// seconds: 2h=7200 24h=86400
// ---------------------------------------------------------
// column generation process
// ---------------------------------------------------------
DecimalFormat df = new DecimalFormat("#0000.00");
oncemore = true;
double[] prevobj = new double[100];
int nbroute;
int previ = -1;
while (oncemore) {
oncemore = false;
// ---------------------------------------s------------------
// solve the current RMP
// ---------------------------------------------------------
if (!cplex.solve()) {
System.out.println("CG: relaxation infeasible!");
return 1E10;
}
prevobj[(++previ) % 100] = cplex.getObjValue();
// store the 30 last obj values to check stability afterwards
// System.out.println(cplex.getStatus());
// cplex.exportModel("model.lp");
// ---------------------------------------------------------
// solve the subproblem to find new columns (if any)
// ---------------------------------------------------------
// first define the new costs for the subproblem objective function
// (SPPRC)
pi = cplex.getDuals(lpmatrix);
for (i = 1; i < userParam.nbclients + 1; i++)
for (j = 0; j < userParam.nbclients + 2; j++)
userParam.cost[i][j] = userParam.dist[i][j] - pi[i - 1];
// start dynamic programming
SPPRC sp = new SPPRC();
ArrayList<route> routesSPPRC = new ArrayList<route>();
nbroute = userParam.nbclients; // arbitrarily limit to the 5 first
// shortest paths with negative cost
// if ((previ>100) &&
// (prevobj[(previ-3)%100]-prevobj[previ%100]<0.0003*Math.abs((prevobj[(previ-99)%100]-prevobj[previ%100]))))
// {
// System.out.print("/");
// complete=true; // it the convergence is too slow, start a "complete"
// shortestpast
// }
sp.shortestPath(userParam, routesSPPRC, nbroute);
sp = null;
// /////////////////////////////
// parameter here
if (routesSPPRC.size() > 0) {
for (route r : routesSPPRC) {
// if (userParam.debug) {
// System.out.println(" "+r.getcost());
// }
ArrayList<Integer> rout = r.getpath();
prevcity = rout.get(1);
cost = userParam.dist[0][prevcity];
IloColumn column = cplex.column(lpmatrix[rout.get(1) - 1], 1.0);
for (i = 2; i < rout.size() - 1; i++) {
city = rout.get(i);
cost += userParam.dist[prevcity][city];
prevcity = city;
column = column.and(cplex.column(lpmatrix[rout.get(i) - 1], 1.0));
// coefficient of y_i in (3.23) => 0 for the other y_p
}
cost += userParam.dist[prevcity][userParam.nbclients + 1];
column = column.and(cplex.column(objfunc, cost));
y.add(cplex.numVar(column, 0.0, Double.MAX_VALUE,
"P" + routes.size())); // creation of the variable y_i
r.setcost(cost);
routes.add(r);
oncemore = true;
}
System.out.print("
CG Iter " + previ + " Current cost: "
+ df.format(prevobj[previ % 100]) + " " + routes.size()
+ " routes");
System.out.flush();
}
//if (previ % 50 == 0)
routesSPPRC = null;
}
System.out.println();
for (i = 0; i < y.getSize(); i++)
routes.get(i).setQ(cplex.getValue(y.getElement(i)));
obj = cplex.getObjValue(); // mmmmhhh: to check. To be entirely safe, we
// should recompute the obj using the distBase
// matrix instead of the dist matrix
cplex.end();
return obj;
} catch (IloException e) {
System.err.println("Concert exception caught '" + e + "' caught");
}
return 1E10;
}
ESPPRC的算法如下，采用label setting算法，感觉速度还可以，具体原理参照往期推文：
public void shortestPath(paramsVRP userParamArg, ArrayList<route> routes,int nbroute) {
label current;
int i,j,idx,nbsol,maxsol;
double d,d2;
int[] checkDom;
float tt,tt2;
Integer currentidx;
this.userParam=userParamArg;
// unprocessed labels list => ordered TreeSet List (?optimal: need to be sorted like this?)
TreeSet<Integer> U = new TreeSet<Integer>(new MyLabelComparator()); // unprocessed labels list
// processed labels list => ordered TreeSet List
TreeSet<Integer> P = new TreeSet<Integer>(new MyLabelComparator()); // unprocessed labels list
// array of labels
labels = new ArrayList<label>(2*userParam.nbclients); // initial size at least larger than nb clients
boolean[] cust= new boolean[userParam.nbclients+2];
cust[0]=true;
for (i=1;i<userParam.nbclients+2;i++)
cust[i]=false;
labels.add(new label(0,-1,0.0,0,0,false,cust)); // first label: start from depot (client 0)
U.add(0);
// for each city, an array with the index of the corresponding labels (for dominance)
checkDom = new int[userParam.nbclients+2];
ArrayList<Integer>[] city2labels = new ArrayList[userParam.nbclients+2];
for (i=0;i<userParam.nbclients+2;i++) {
city2labels[i]=new ArrayList<Integer>();
checkDom[i]=0; // index of the first label in city2labels that needs to be checked for dominance (last labels added)
}
city2labels[0].add(0);
nbsol = 0;
maxsol = 2 * nbroute;
while ((U.size()>0) && (nbsol<maxsol)) {
// second term if we want to limit to the first solutions encountered to speed up the SPPRC (perhaps not the BP)
// remark: we'll keep only nbroute, but we compute 2xnbroute! It makes a huge difference=>we'll keep the most negative ones
// this is something to analyze further! how many solutions to keep and which ones?
// process one label => get the index AND remove it from U
currentidx = U.pollFirst();
current = labels.get(currentidx);
// check for dominance
// code not fully optimized:
int l1,l2;
boolean pathdom;
label la1,la2;
ArrayList<Integer> cleaning = new ArrayList<Integer>();
for (i = checkDom[current.city]; i < city2labels[current.city].size(); i++) { // check for dominance between the labels added since the last time we came here with this city and all the other ones
for (j = 0; j < i; j++) {
l1 = city2labels[current.city].get(i);
l2 = city2labels[current.city].get(j);
la1 = labels.get(l1);
la2 = labels.get(l2);
if (!(la1.dominated || la2.dominated)) { // could happen since we clean 'city2labels' thanks to 'cleaning' only after the double loop
pathdom = true;
for (int k = 1; pathdom && (k < userParam.nbclients+2); k++)
pathdom=(!la1.vertexVisited[k] || la2.vertexVisited[k]);
if (pathdom && (la1.cost<=la2.cost) && (la1.ttime<=la2.ttime) && (la1.demand<=la2.demand)) {
labels.get(l2).dominated = true;
U.remove((Integer) l2);
cleaning.add(l2);
pathdom = false;
//System.out.print(" ###Remove"+l2);
}
pathdom = true;
for (int k = 1; pathdom && (k < userParam.nbclients + 2); k++)
pathdom = (!la2.vertexVisited[k] || la1.vertexVisited[k]);
if (pathdom && (la2.cost<=la1.cost) && (la2.ttime<=la1.ttime) && (la2.demand<=la1.demand)) {
labels.get(l1).dominated = true;
U.remove(l1);
cleaning.add(l1);
//System.out.print(" ###Remove"+l1);
j = city2labels[current.city].size();
}
}
}
}
for (Integer c : cleaning)
city2labels[current.city].remove((Integer) c); // a little bit confusing but ok since c is an Integer and not an int!
cleaning = null;
checkDom[current.city] = city2labels[current.city].size(); // update checkDom: all labels currently in city2labels were checked for dom.
// expand REF
if (!current.dominated){
//System.out.println("Label "+current.city+" "+current.indexPrevLabel+" "+current.cost+" "+current.ttime+" "+current.dominated);
if (current.city == userParam.nbclients + 1) { // shortest path candidate to the depot!
if (current.cost<-1e-7) { // SP candidate for the column generation
P.add(currentidx);
nbsol=0;
for (Integer labi : P) {
label s = labels.get(labi);
if (!s.dominated)
nbsol++;
}
}
} else { // if not the depot, we can consider extensions of the path
for (i = 0; i < userParam.nbclients + 2; i++) {
if ((!current.vertexVisited[i]) && (userParam.dist[current.city][i] < userParam.verybig-1e-6)) { // don't go back to a vertex already visited or along a forbidden edge
// ttime
tt = (float) (current.ttime + userParam.ttime[current.city][i] + userParam.s[current.city]);
if (tt < userParam.a[i])
tt = userParam.a[i];
// demand
d = current.demand + userParam.d[i];
//System.out.println(" -- "+i+" d:"+d+" t:"+tt);
// is feasible?
if ((tt <= userParam.b[i]) && (d <= userParam.capacity)) {
idx = labels.size();
boolean[] newcust = new boolean[userParam.nbclients + 2];
System.arraycopy(current.vertexVisited, 0, newcust, 0, userParam.nbclients + 2);
newcust[i] = true;
//speedup: third technique - Feillet 2004 as mentioned in Laporte's paper
for (j=1;j<=userParam.nbclients;j++)
if (!newcust[j]) {
tt2=(float) (tt+userParam.ttime[i][j]+userParam.s[i]);
d2=d+userParam.d[j];
if ((tt2>userParam.b[j]) || (d2>userParam.capacity))
newcust[j]=true; // useless to visit this client
}
labels.add(new label(i, currentidx, current.cost+userParam.cost[current.city][i], tt, d, false, newcust)); // first label: start from depot (client 0)
if (!U.add((Integer) idx)) {
// only happens if there exists already a label at this vertex with the same cost, time and demand and visiting the same cities before
// It can happen with some paths where the order of the cities is permuted
labels.get(idx).dominated = true; // => we can forget this label and keep only the other one
} else
city2labels[i].add(idx);
}
}
}
}
}
}
// clean
checkDom = null;
// filtering: find the path from depot to the destination
Integer lab;
i = 0;
while ((i < nbroute) && ((lab = P.pollFirst()) != null)) {
label s = labels.get(lab);
if (!s.dominated) {
if (/*(i < nbroute / 2) ||*/ (s.cost < -1e-4)) {
// System.out.println(s.cost);
// if(s.cost > 0) {
// System.out.println("warning >>>>>>>>>>>>>>>>>>>>");
// }
route newroute = new route();
newroute.setcost(s.cost);
newroute.addcity(s.city);
int path = s.indexPrevLabel;
while (path >= 0) {
newroute.addcity(labels.get(path).city);
path = labels.get(path).indexPrevLabel;
}
newroute.switchpath();
routes.add(newroute);
i++;
}
}
}
}
没想到吧，一个算法就包含了这么多知识点，没点基础真的是搞不定的哦~
大家好好加油吧哈哈。
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