题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3491
思路：由于每个城市顶点都具有权值,所以对于每个城市拆成两个点u和所对应的u',之间连容量为w的边,S,H两点不会算在最小割中,所以将这两点拆点,拆点后容量为无穷,添加源点vs(0)和汇点vt(2*n+1),加边(vs,s,INF)和(t+n,vt,INF),对于两相连的城市,用其中一个点的第二个点去连另一个点的第一个点，边流量为无穷大,然后求解最大流即可.
View Code
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 #define MAXN 222 6 #define MAXM 222222 7 #define inf 1<<30 8 struct Edge{ 9 int v,cap,next; 10 }edge[MAXM]; 11 12 int head[MAXN]; 13 int pre[MAXN]; 14 int cur[MAXN]; 15 int level[MAXN]; 16 int gap[MAXN]; 17 int NV,NE,n,m,vs,vt; 18 19 void Insert(int u,int v,int cap,int cc=0){ 20 edge[NE].v=v;edge[NE].cap=cap; 21 edge[NE].next=head[u];head[u]=NE++; 22 23 edge[NE].v=u;edge[NE].cap=cc; 24 edge[NE].next=head[v];head[v]=NE++; 25 } 26 27 28 int SAP(int vs,int vt){ 29 memset(pre,-1,sizeof(pre)); 30 memset(level,0,sizeof(level)); 31 memset(gap,0,sizeof(gap)); 32 for(int i=1;i<=NV;i++)cur[i]=head[i]; 33 int u=pre[vs]=vs,maxflow=0,aug=-1; 34 gap[0]=NV; 35 while(level[vs]<NV){ 36 loop: 37 for(int &i=cur[u];i!=-1;i=edge[i].next){ 38 int v=edge[i].v; 39 if(edge[i].cap&&level[u]==level[v]+1){ 40 aug==-1?aug=edge[i].cap:aug=min(aug,edge[i].cap); 41 pre[v]=u; 42 u=v; 43 if(v==vt){ 44 maxflow+=aug; 45 for(u=pre[u];v!=vs;v=u,u=pre[u]){ 46 edge[cur[u]].cap-=aug; 47 edge[cur[u]^1].cap+=aug; 48 } 49 aug=-1; 50 } 51 goto loop; 52 } 53 } 54 int minlevel=NV; 55 for(int i=head[u];i!=-1;i=edge[i].next){ 56 int v=edge[i].v; 57 if(edge[i].cap&&minlevel>level[v]){ 58 cur[u]=i; 59 minlevel=level[v]; 60 } 61 } 62 gap[level[u]]--; 63 if(gap[level[u]]==0)break; 64 level[u]=minlevel+1; 65 gap[level[u]]++; 66 u=pre[u]; 67 } 68 return maxflow; 69 } 70 71 int main(){ 72 int _case,num,u,v,s,t; 73 scanf("%d",&_case); 74 while(_case--){ 75 scanf("%d%d%d%d",&n,&m,&s,&t); 76 NE=0,NV=2*n+2; 77 vs=0,vt=2*n+1; 78 memset(head,-1,sizeof(head)); 79 Insert(vs,s,inf); 80 Insert(t+n,vt,inf); 81 for(int i=1;i<=n;i++){ 82 scanf("%d",&num); 83 if(i!=s&&i!=t){Insert(i,i+n,num);} 84 else Insert(i,i+n,inf); 85 } 86 for(int i=1;i<=m;i++){ 87 scanf("%d%d",&u,&v); 88 Insert(u+n,v,inf); 89 Insert(v+n,u,inf); 90 } 91 printf("%d\n",SAP(vs,vt)); 92 } 93 return 0; 94 }