• POJ-1655 Balancing Act


    题目大意:一棵n个节点的树,找出最大子树最小的节点。

    题目分析:过程类似求重心。

    代码如下:

    # include<iostream>
    # include<cstdio>
    # include<cstring>
    # include<vector>
    # include<queue>
    # include<list>
    # include<set>
    # include<map>
    # include<string>
    # include<cmath>
    # include<cstdlib>
    # include<algorithm>
    using namespace std;
    # define LL long long
    
    const int N=1005;
    const int INF=1000000000;
    
    int n,dp[N*20];
    int size[N*20];
    vector<int>e[N*20];
    
    void init()
    {
    	scanf("%d",&n);
    	for(int i=1;i<=n;++i)
    		e[i].clear();
    	int a,b;
    	for(int i=1;i<n;++i){
    		scanf("%d%d",&a,&b);
    		e[a].push_back(b);
    		e[b].push_back(a);
    	}
    }
    
    void dfs(int u,int fa)
    {
    	size[u]=1;
    	for(int i=0;i<e[u].size();++i){
    		int v=e[u][i];
    		if(v==fa) continue;
    		dfs(v,u);
    		size[u]+=size[v];
    	}
    }
    
    void dfs1(int u,int fa)
    {
    	dp[u]=n-size[u];
    	for(int i=0;i<e[u].size();++i){
    		int v=e[u][i];
    		if(v==fa) continue;
    		dp[u]=max(dp[u],size[v]);
    		dfs1(v,u);
    	}
    }
    
    void solve()
    {
    	dfs(1,-1);
    	dfs1(1,-1);
    	dp[0]=INF;
    	int root=0;
    	for(int i=1;i<=n;++i)
    		if(dp[i]<dp[root])
    			root=i;
    	printf("%d %d
    ",root,dp[root]);
    }
    
    int main()
    {
    	int T;
    	scanf("%d",&T);
    	while(T--)
    	{
    		init();
    		solve();
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/20143605--pcx/p/5399818.html
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