hdu3452 無向樹去掉最小的邊集使任何葉子與根不連通 / 最小割

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思路一下就上來了，葉子向匯點連邊，inf保證不會成為割，跑根到匯點最小割即可。注意無向樹雙向建邊。基礎題，分分鐘1A:

#include<iostream>#include<queue>#include<cstdio>#include<cstring>#include<set>#include<vector>using namespace std;const int inf=0x3f3f3f3f;const int maxv=1005,maxe=10000;int nume=0;int head[maxv];int e[maxe][3];void inline adde(int i,int j,int c){    e[nume][0]=j;e[nume][1]=head[i];head[i]=nume;    e[nume++][2]=c;    e[nume][0]=i;e[nume][1]=head[j];head[j]=nume;    e[nume++][2]=0;}int ss,tt,n,m;int vis[maxv];int lev[maxv];bool bfs(){    for(int i=0;i<maxv;i++)      vis[i]=lev[i]=0;    queue<int>q;    q.push(ss);    vis[ss]=1;    while(!q.empty())    {        int cur=q.front();        q.pop();        for(int i=head[cur];i!=-1;i=e[i][1])        {            int v=e[i][0];            if(!vis[v]&&e[i][2]>0)            {                lev[v]=lev[cur]+1;                vis[v]=1;                q.push(v);            }        }    }    return vis[tt];}int dfs(int u,int minf){    if(u==tt||minf==0)return minf;    int sumf=0,f;    for(int i=head[u];i!=-1&&minf;i=e[i][1])    {        int v=e[i][0];        if(lev[v]==lev[u]+1&&e[i][2]>0)        {            f=dfs(v,minf<e[i][2]?minf:e[i][2]);            e[i][2]-=f;e[i^1][2]+=f;            sumf+=f;minf-=f;        }    }    if(!sumf) lev[u]=-1;    return sumf;}int dinic(){    int sum=0;    while(bfs())sum+=dfs(ss,inf);    return sum;};int ind[maxv];void read_build(){    int aa,bb,cc;    for(int i=0;i<n-1;i++)     {        scanf("%d%d%d",&aa,&bb,&cc);        adde(aa,bb,cc);        adde(bb,aa,cc);        ind[aa]++;ind[bb]++;     }     for(int i=1;i<=n;i++)        if(i!=m&&ind[i]==1)        {            adde(i,tt,inf);        }  /*  for(int i=0;i<=tt;i++)      for(int j=head[i];j!=-1;j=e[j][1])      {          if(j%2==0)          printf("%d->%d:%d\n",i,e[j][0],e[j][2]);      }*/}void init(){    nume=0;    ss=m;tt=n+1;    for(int i=0;i<=tt;i++)      {          head[i]=-1;ind[i]=0;      }}int main(){      while(~scanf("%d%d",&n,&m)&&(n||m))     {         init();        read_build();       int ans;       ans=dinic();      printf("%d\n",ans);    }    return 0;}