poj 1655 树形dp

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题意相当于给你一棵树让你求每个点的不同子树上的节点个数吧注意存边的时候存双向边

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;struct node{int to,next;}A[40010];int tot,list[20010],mark[20010],n;int add(int a,int b){A[++tot].to = b;               A[tot].next = list[a];    list[a] = tot;  return 0; }int dp[20010],subtree[20010];int max(int a,int b){return a>b?a:b;}int dfs(int point){mark[point]=1;int b;int k=list[point];if(k==-1) {dp[point]=n-1;subtree[point]=0;return 0;}for(;k!=-1;k = A[k].next)    {   b = A[k].to;   if(mark[b]) continue;   dfs(b);   subtree[point]+=subtree[b]+1;   dp[point]=max(dp[point],subtree[b]+1);   }   dp[point]=max(dp[point],n-subtree[point]-1);   return 0;}int main(){int i,j,T;scanf("%d",&T);while(T--){scanf("%d",&n);tot=1;memset(list,-1,sizeof(list));memset(mark,0,sizeof(mark));for(i=1;i<n;i++){int a,b;scanf("%d%d",&a,&b);add(a,b);add(b,a);}memset(dp,0,sizeof(dp));memset(subtree,0,sizeof(subtree));memset(mark,0,sizeof(mark));dfs(1);int Min=3000000,k;for(i=1;i<=n;i++)if(dp[i]<Min){Min=dp[i];k=i;}printf("%d %d\n",k,Min);}return 0;}

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