题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=569&problem=4277&mosmsg=Submission+received+with+ID+2182664
题目大意:给定一个带权有向图(v,e)，求两条互不相交的从1到v的路径(不经过同一个点)，使两条路径的权和最小.
分析:即求从1到v的最小费用流,到v的流量为2,且每个点只能经过一次.加一个从v出发的点v',费用为0,容量为2,求1到v'的最小费用流即满足前一条件;将2到v-1每个点拆分成两个点,费用为0,容量为1,再求从1到v'的最小费用流即可.
一定要注意求mcmf时设置的INF要恰当!!!不然就溢出然后疯狂WA了!!!
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<vector> 5 #include<queue> 6 using namespace std; 7 const int maxn=2005,maxm=20005,INF=1000000; 8 struct Edge{ 9 int from,to,cap,flow,cost; 10 Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w){} 11 }; 12 13 struct MCMF{ 14 int n,m; 15 vector<Edge> edges; 16 vector<int> G[maxn]; 17 int inq[maxn],d[maxn],p[maxn],a[maxn]; 18 19 void init(int n){ 20 this->n=n; 21 this->m=0; 22 for(int i=0;i<n;i++)G[i].clear(); 23 edges.clear(); 24 } 25 26 void AddEdge(int from,int to,int cap,int cost){ 27 edges.push_back(Edge(from,to,cap,0,cost)); 28 edges.push_back(Edge(to,from,0,0,-cost)); 29 G[from].push_back(this->m++); 30 G[to].push_back(this->m++); 31 } 32 33 bool BellmanFord(int s,int t,int &flow,long long &cost){ 34 for(int i=0;i<n;i++)d[i]=INF; 35 memset(inq,0,sizeof(inq)); 36 d[s]=0;inq[s]=1;p[s]=0;a[s]=INF; 37 queue<int> Q; 38 Q.push(s); 39 while(!Q.empty()){ 40 int u=Q.front();Q.pop(); 41 inq[u]=0; 42 43 for(int i=0;i<G[u].size();i++){ 44 Edge &e=edges[G[u][i]]; 45 if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){ 46 d[e.to]=d[u]+e.cost; 47 p[e.to]=G[u][i]; 48 a[e.to]=min(a[u],e.cap-e.flow); 49 if(!inq[e.to]){Q.push(e.to);inq[e.to]=1;} 50 } 51 } 52 } 53 if(d[t]==INF)return false; 54 flow+=a[t]; 55 cost+=(long long)d[t]*(long long)a[t]; 56 for(int u=t;u!=s;u=edges[p[u]].from){ 57 edges[p[u]].flow+=a[t]; 58 edges[p[u]^1].flow-=a[t]; 59 } 60 return true; 61 } 62 63 int Mincost(int s,int t,long long &cost){ 64 int flow=0;cost=0; 65 while(BellmanFord(s,t,flow,cost)); 66 return flow; 67 } 68 }; 69 70 int main(){ 71 //freopen("e:\in.txt","r",stdin); 72 int n,m,a,b,c; 73 MCMF mcmf; 74 mcmf.init(2*n); 75 for(int i=1;i<n-1;i++){ 76 mcmf.AddEdge(i,i+n,1,0); 77 } 78 for(int i=0;i<m;i++){ 79 cin>>a>>b>>c; 80 a--;b--; 81 if(a==0||a==n-1){ 82 mcmf.AddEdge(a,b,1,c); 83 } 84 else{ 85 mcmf.AddEdge(a+n,b,1,c); 86 } 87 } 88 mcmf.AddEdge(n-1,2*n-1,2,0); 89 long long cost=0; 90 mcmf.Mincost(0,2*n-1,cost); 91 cout<<cost<<endl; 92 return 0; 93 }