NYOJ 679 The Weight of Tree 搜索+dp+邻接表

最新推荐文章于 2018-05-19 21:31:25 发布

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本文介绍了一种解决树形结构中寻找最大权值子树的问题的方法。通过深度优先搜索算法，实现对树形结构的有效遍历，并在回溯过程中更新节点权值，确保最终找到的子树具有最大的节点权值之和。

The Weight of Tree

时间限制：3000 ms | 内存限制：65535 KB

难度：4

描述

456 has a tree of n nodes, each node is assigned with an integer number. Now 456 wants to select a subtree, such that the sum of all integers on the nodes of the subtree is maxmized. Can you help him?

输入

On the first line of the input is an integer T, and then T cases follows. Each case begins with a positive integer n(1 <= n <= 10^5), then n numbers Wi(-1000 <= Wi <= 1000),Wi for the number on the ith node. Then n - 1 lines follows, each line contains two numbers a, b(1 <= a, b <= n)indicate that there is a edge between node a and b.

输出

For each test case, output one integer on a line, the maximized sum can be achieved by selecting a subtree.

样例输入

3
1
5
2
5 -5
1 2
5
-2 -3 7 -1 4
1 2
2 3
3 4
2 5

样例输出

5
5
8

从第一个点一直往下深搜，然后回溯，判断子节点的值是否大于0，如果大于0，父节点的值变为当前的值加上子节点的值。最后输出最大值即可。

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
#define N 100005
using namespace std;
vector<int> vec[N];
int vis[N],dp[N],MAX;
void dfs(int x)
{
	vis[x]=1;
	for(int i=0;i<vec[x].size();i++)
	{
		int y=vec[x][i];
		if(vis[y])
		   continue;
        dfs(y);
        if(dp[y]>0)
           dp[x]+=dp[y];
		MAX=max(dp[x],MAX);
    }
}
int main()
{
	int t,n,i,a,b;
	scanf("%d",&t);
	while(t--)
	{
		memset(vis,0,sizeof(vis));
		memset(vec,0,sizeof(vec));
		scanf("%d",&n);
		MAX=-999999;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&dp[i]);
			MAX=max(dp[i],MAX);
		}
		for(i=1;i<n;i++)
		{
			scanf("%d%d",&a,&b);
			vec[a].push_back(b);
			vec[b].push_back(a);
		}
		dfs(1);
		printf("%d\n",MAX);
	}
	return 0;
}