Coding Contest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2653 Accepted Submission(s): 579
Problem Description
A coding contest will be held in this university, in a huge playground. The whole playground would be divided into N blocks, and there would be M directed paths linking these blocks. The i-th path goes from the ui-th block to the vi-th block. Your task is to solve the lunch issue. According to the arrangement, there are si competitors in the i-th block. Limited to the size of table, bi bags of lunch including breads, sausages and milk would be put in the i-th block. As a result, some competitors need to move to another block to access lunch. However, the playground is temporary, as a result there would be so many wires on the path.
For the i-th path, the wires have been stabilized at first and the first competitor who walker through it would not break the wires. Since then, however, when a person go through the i - th path, there is a chance of pi to touch
the wires and affect the whole networks. Moreover, to protect these wires, no more than ci competitors are allowed to walk through the i-th path.
Now you need to find a way for all competitors to get their lunch, and minimize the possibility of network crashing.
Input
The first line of input contains an integer t which is the number of test cases. Then t test cases follow.
For each test case, the first line consists of two integers N (N ≤ 100) and M (M ≤ 5000). Each of the next N lines contains two integers si and bi (si , bi ≤ 200).
Each of the next M lines contains three integers ui , vi and ci(ci ≤ 100) and a float-point number pi(0 < pi < 1).
It is guaranteed that there is at least one way to let every competitor has lunch.
Output
For each turn of each case, output the minimum possibility that the networks would break down. Round it to 2 digits.
Sample Input
1 4 4 2 0 0 3 3 0 0 3 1 2 5 0.5 3 2 5 0.5 1 4 5 0.5 3 4 5 0.5
Sample Output
0.50
思路：
这是一个最小费用最大流的变形题目。
要仔细读懂题目，题目要求的是崩溃的最小可能性。由于崩溃可能是由一条边或多条边一起的，所以正着求崩溃概率很难求，我们反着思考，用1去减每条边都不引起图崩溃的概率即为崩溃的概率。
由于求的是最小，而且边有容量限制，我们很自然的想到了最小费用最大流（对于不崩溃的概率应该是最大费用最大流）
建图时，某个点有人则想s连一条容量为人数的边，有食物则向t连一条容量为是无数的边，其他的边按照题目的数据相连即可，这个很好处理。
还有一个条件是某条路第一遍走的时候是没有花费的。我的处理方法是将一条容量为c，花费为w的边拆成两条边：容量为c-1，花费为w的一条边；容量为1，花费为1.0的一条边（花费为1.0是为了确保不会对最终结果产生影响）。
细节很多，由于概率要想乘所以初始化时要注意，算一条增广路径对总结果的贡献时也要注意是乘方而不是简单的相乘。
代码1：求增广路径时直接求最长路
View Code1 /* 2 * @FileName: D:代码与算法2017训练比赛2016青岛区域赛g.cpp 3 * @Author: Pic 4 * @Created Time: 2017/10/7 12:46:48 5 */ 6 #include <bits/stdc++.h> 7 using namespace std; 8 const double INF = -1*10.0; 9 const int maxn=200+10; 10 11 struct Edge 12 { 13 int from,to,cap,flow; 14 double cost; 15 Edge(){} 16 Edge(int f,int t,int c,int fl,double co):from(f),to(t),cap(c),flow(fl),cost(co){} 17 }; 18 19 struct MCMF 20 { 21 int n,m,s,t; 22 vector<Edge> edges; 23 vector<int> G[maxn]; 24 bool inq[maxn]; //是否在队列 25 double d[maxn]; //Bellman_ford单源最短路径 26 int p[maxn]; //p[i]表从s到i的最小费用路径上的最后一条弧编号 27 int a[maxn]; //a[i]表示从s到i的最小残量 28 29 //初始化 30 void init(int n,int s,int t) 31 { 32 this->n=n, this->s=s, this->t=t; 33 edges.clear(); 34 for(int i=0;i<n;++i) G[i].clear(); 35 } 36 37 //添加一条有向边 38 void AddEdge(int from,int to,int cap,double cost) 39 { 40 edges.push_back(Edge(from,to,cap,0,cost)); 41 edges.push_back(Edge(to,from,0,0,-cost)); 42 m=edges.size(); 43 G[from].push_back(m-2); 44 G[to].push_back(m-1); 45 } 46 47 //求一次增广路 48 bool BellmanFord(int &flow, double &cost) 49 { 50 for(int i=0;i<n;++i) d[i]=INF; 51 memset(inq,0,sizeof(inq)); 52 d[s]=1.0, a[s]=1e9+30, inq[s]=true, p[s]=0; 53 queue<int> Q; 54 Q.push(s); 55 while(!Q.empty()) 56 { 57 //cout<<"Q"<<endl; 58 int u=Q.front(); Q.pop(); 59 inq[u]=false; 60 for(int i=0;i<G[u].size();++i) 61 { 62 Edge &e=edges[G[u][i]]; 63 double tmp=(d[u]*e.cost); 64 //tmp=-1*fabs(tmp); 65 if( e.cap>e.flow && d[e.to] < tmp ) 66 { 67 d[e.to]=tmp; 68 p[e.to]=G[u][i]; 69 a[e.to]= min(a[u],e.cap-e.flow); 70 if(!inq[e.to]){ Q.push(e.to); inq[e.to]=true; } 71 } 72 } 73 } 74 //cout<<d[t]<<endl; 75 //if(fabs(d[t]-INF)<1e-3) return false; 76 if(d[t]<0) return false; 77 flow += a[t]; 78 for(int i=0;i<a[t];i++) cost*=(d[t]); 79 int u=t; 80 while(u!=s) 81 { 82 edges[p[u]].flow += a[t]; 83 edges[p[u]^1].flow -=a[t]; 84 u = edges[p[u]].from; 85 } 86 return true; 87 } 88 89 //求出最小费用最大流 90 double Min_cost() 91 { 92 int flow=0;double cost=1.0; 93 while(BellmanFord(flow,cost)) ; 94 return cost; 95 } 96 }MM; 97 int main() 98 { 99 // freopen("data.in","r",stdin); 100 //freopen("data.out","w",stdout); 101 int t; 102 scanf("%d",&t); 103 while(t--){ 104 int n,m; 105 scanf("%d%d",&n,&m); 106 MM.init(n+2,0,n+1); 107 int x,y; 108 for(int i=1;i<=n;i++){ 109 scanf("%d%d",&x,&y); 110 if(x!=0) 111 MM.AddEdge(0,i,x,1.0); 112 if(y!=0) 113 MM.AddEdge(i,n+1,y,1.0); 114 } 115 int u,v,c;double w; 116 for(int i=1;i<=m;i++){ 117 scanf("%d%d%d%lf",&u,&v,&c,&w); 118 //MM.AddEdge(u,v,c,w); 119 if(c>0){ 120 MM.AddEdge(u,v,c-1,1.0-w); 121 MM.AddEdge(u,v,1,1.0); 122 } 123 } 124 double res=fabs(MM.Min_cost()); 125 //cout<<res<<endl; 126 res=1.0-res; 127 if(res>=1.0) res=1.0; 128 printf("%.2lf ",res); 129 } 130 return 0; 131 }代码2：花费取负，求最小费用最大流，注意初始化的值
View Code1 /* 2 * @FileName: D:代码与算法2017训练比赛2016青岛区域赛g.cpp 3 * @Author: Pic 4 * @Created Time: 2017/10/7 12:46:48 5 */ 6 #include <bits/stdc++.h> 7 using namespace std; 8 const double INF = 1e8; 9 const int maxn=200+10; 10 11 struct Edge 12 { 13 int from,to,cap,flow; 14 double cost; 15 Edge(){} 16 Edge(int f,int t,int c,int fl,double co):from(f),to(t),cap(c),flow(fl),cost(co){} 17 }; 18 19 struct MCMF 20 { 21 int n,m,s,t; 22 vector<Edge> edges; 23 vector<int> G[maxn]; 24 bool inq[maxn]; //是否在队列 25 double d[maxn]; //Bellman_ford单源最短路径 26 int p[maxn]; //p[i]表从s到i的最小费用路径上的最后一条弧编号 27 int a[maxn]; //a[i]表示从s到i的最小残量 28 29 //初始化 30 void init(int n,int s,int t) 31 { 32 this->n=n, this->s=s, this->t=t; 33 edges.clear(); 34 for(int i=0;i<n;++i) G[i].clear(); 35 } 36 37 //添加一条有向边 38 void AddEdge(int from,int to,int cap,double cost) 39 { 40 edges.push_back(Edge(from,to,cap,0,cost)); 41 edges.push_back(Edge(to,from,0,0,-cost)); 42 m=edges.size(); 43 G[from].push_back(m-2); 44 G[to].push_back(m-1); 45 } 46 47 //求一次增广路 48 bool BellmanFord(int &flow, double &cost) 49 { 50 for(int i=0;i<n;++i) d[i]=INF; 51 memset(inq,0,sizeof(inq)); 52 d[s]=1.0, a[s]=INF, inq[s]=true, p[s]=0; 53 queue<int> Q; 54 Q.push(s); 55 while(!Q.empty()) 56 { 57 // cout<<"Q"<<endl; 58 int u=Q.front(); Q.pop(); 59 inq[u]=false; 60 for(int i=0;i<G[u].size();++i) 61 { 62 Edge &e=edges[G[u][i]]; 63 double tmp=(d[u]*e.cost); 64 tmp=-1*fabs(tmp); 65 if( e.cap>e.flow && d[e.to] > tmp ) 66 { 67 d[e.to]=tmp; 68 p[e.to]=G[u][i]; 69 a[e.to]= min(a[u],e.cap-e.flow); 70 if(!inq[e.to]){ Q.push(e.to); inq[e.to]=true; } 71 } 72 } 73 } 74 if(fabs(d[t]-INF)<1e3) return false; 75 flow += a[t]; 76 for(int i=0;i<a[t];i++) cost*=(d[t]); 77 int u=t; 78 while(u!=s) 79 { 80 edges[p[u]].flow += a[t]; 81 edges[p[u]^1].flow -=a[t]; 82 u = edges[p[u]].from; 83 } 84 return true; 85 } 86 87 //求出最小费用最大流 88 double Min_cost() 89 { 90 int flow=0;double cost=1.0; 91 while(BellmanFord(flow,cost)); 92 return cost; 93 } 94 }MM; 95 int main() 96 { 97 // freopen("data.in","r",stdin); 98 //freopen("data.out","w",stdout); 99 int t; 100 scanf("%d",&t); 101 while(t--){ 102 int n,m; 103 scanf("%d%d",&n,&m); 104 MM.init(n+2,0,n+1); 105 int x,y; 106 for(int i=1;i<=n;i++){ 107 scanf("%d%d",&x,&y); 108 if(x!=0) 109 MM.AddEdge(0,i,x,1.0); 110 if(y!=0) 111 MM.AddEdge(i,n+1,y,1.0); 112 } 113 int u,v,c;double w; 114 for(int i=1;i<=m;i++){ 115 scanf("%d%d%d%lf",&u,&v,&c,&w); 116 //MM.AddEdge(u,v,c,w); 117 if(c>0){ 118 MM.AddEdge(u,v,c-1,w-1.0); 119 MM.AddEdge(u,v,1,1.0); 120 } 121 } 122 double res=fabs(MM.Min_cost()); 123 //cout<<res<<endl; 124 res=1.0-res; 125 if(res>=1.0) res=1.0; 126 printf("%.2lf ",res); 127 } 128 return 0; 129 }
