hdu 4276

最新推荐文章于 2024-10-01 18:52:56 发布

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本文介绍了一段使用C++解决复杂路径问题的代码，包括路径查找、路径优化等核心功能。通过深入分析代码逻辑，读者可以了解如何在C++中高效处理路径相关任务。

/*
 * p4276.cpp
 *
 *  Created on: 2013-7-24
 *      Author: zy
 */
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<algorithm>
using namespace std;
int dp[110][510], value[110];
vector<pair<int, int> > son[110];
int n, m;
int time1;
void init()
{
	memset(dp, 0, sizeof(dp));
	for (int i = 0; i <= n; i++)
		son[i].clear();
}
bool dfs1(int u, int pa)
{
	if (u == n)
		return true;
	for (int i = 0; i < son[u].size(); i++)
	{
		int v = son[u][i].first;
		if (v == pa)
			continue;
		if (dfs1(v, u))
		{
			time1 += son[u][i].second;
			son[u][i].second = 0;
			return true;
		}
	}
	return false;
}
void dfs2(int u, int pa)
{
	for (int i = 0; i <= m; i++)
		dp[u][i] = value[u];
	for (int i = 0; i < son[u].size(); i++)
	{
		int v = son[u][i].first;
		int l = son[u][i].second;
		if (v == pa)
			continue;
		dfs2(v, u);
		for (int j = m; j >= 0; j--)
			for (int k = 0; k <= j; k++)
				if (j - k - 2 * l >= 0)
					dp[u][j] = max(dp[u][j], dp[u][j - k - 2 * l] + dp[v][k]);

	}
}
int main()
{
	while (scanf("%d%d", &n, &m) != EOF)
	{
		init();
		int x, y, t;
		for (int i = 1; i < n; i++)
		{
			scanf("%d%d%d", &x, &y, &t);
			son[x].push_back(make_pair(y, t));
			son[y].push_back(make_pair(x, t));
		}
		for (int i = 1; i <= n; i++)
			scanf("%d", &value[i]);
		time1 = 0;
		dfs1(1, -1);
		m -= time1;
		if (m < 0)
		{
			printf(
					"Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");
			continue;
		}
		dfs2(1, -1);
		printf("%d\n", dp[1][m]);
	}
	return 0;
}