题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3452
思路：一颗带边权的树,最小割求树根和叶子结点不连通。可设汇点为n+1,注意只用叶子结点才能和汇点连容量为inf的边。
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1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<set> 6 using namespace std; 7 #define MAXN 2222 8 #define MAXM 222222 9 #define inf 1<<30 10 11 struct Edge{ 12 int v,cap,next; 13 }edge[MAXM]; 14 15 16 int head[MAXN]; 17 int pre[MAXN]; 18 int cur[MAXN]; 19 int level[MAXN]; 20 int gap[MAXN]; 21 int NV,NE,n,m,vs,vt; 22 23 void Insert(int u,int v,int cap,int cc=0){ 24 edge[NE].v=v;edge[NE].cap=cap; 25 edge[NE].next=head[u];head[u]=NE++; 26 27 edge[NE].v=u;edge[NE].cap=cc; 28 edge[NE].next=head[v];head[v]=NE++; 29 } 30 31 32 int SAP(int vs,int vt){ 33 memset(pre,-1,sizeof(pre)); 34 memset(level,0,sizeof(level)); 35 memset(gap,0,sizeof(gap)); 36 for(int i=0;i<=NV;i++)cur[i]=head[i]; 37 int u=pre[vs]=vs,maxflow=0,aug=-1; 38 gap[0]=NV; 39 while(level[vs]<NV){ 40 loop: 41 for(int &i=cur[u];i!=-1;i=edge[i].next){ 42 int v=edge[i].v; 43 if(edge[i].cap&&level[u]==level[v]+1){ 44 aug==-1?aug=edge[i].cap:aug=min(aug,edge[i].cap); 45 pre[v]=u; 46 u=v; 47 if(v==vt){ 48 maxflow+=aug; 49 for(u=pre[u];v!=vs;v=u,u=pre[u]){ 50 edge[cur[u]].cap-=aug; 51 edge[cur[u]^1].cap+=aug; 52 } 53 aug=-1; 54 } 55 goto loop; 56 } 57 } 58 int minlevel=NV; 59 for(int i=head[u];i!=-1;i=edge[i].next){ 60 int v=edge[i].v; 61 if(edge[i].cap&&minlevel>level[v]){ 62 cur[u]=i; 63 minlevel=level[v]; 64 } 65 } 66 if(--gap[level[u]]==0)break; 67 level[u]=minlevel+1; 68 gap[level[u]]++; 69 u=pre[u]; 70 } 71 return maxflow; 72 } 73 74 int main(){ 75 while(~scanf("%d%d",&n,&m)&&n&&m){ 76 vs=m,vt=n+1,NV=vt,NE=0; 77 memset(head,-1,sizeof(head)); 78 set<int>st[MAXN]; 79 for(int i=1;i<=n-1;i++){ 80 int u,v,w; 81 scanf("%d%d%d",&u,&v,&w); 82 Insert(u,v,w); 83 Insert(v,u,w); 84 st[u].insert(v); 85 st[v].insert(u); 86 } 87 //只有叶子结点才与汇点连容量为inf的边 88 for(int i=0;i<=n;i++){ 89 if(st[i].size()==1&&i!=vs)Insert(i,vt,inf); 90 } 91 printf("%d\n",SAP(vs,vt)); 92 } 93 return 0; 94 } 95 96 97 98 99 100 101 102 103 104 105