UVALive - 6800 The Mountain of Gold?（Bellman-ford找负权回路，dfs）

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题目链接

https://icpcarchive.ecs.baylor.edu/external/68/6800.pdf

bellman-ford照模板打了一段，能够找到负权回路，问题就是判断0点在不在负权回路中了，于是写了个记忆化dfs。

#include<iostream>  
#include<cstdio>  
#include<cstring>
#include<string>
#include<vector>
using namespace std;

#define MAX 0x3f3f3f3f  
#define N 10100  
int nodenum, edgenum;  

typedef struct Edge   
{
	int u, v;
	int cost;
}Edge;

Edge edge[N];
int dis[N];

bool vis[N], dp[N];
vector<int> G[N];
bool dfs(int n)
{
	if (n == 0) return true;
	if (vis[n]) return dp[n];
	vis[n] = true;

	for (int i = 0; i < G[n].size(); i++)
	{
		if (dfs(G[n][i]))
		{
			return dp[n] = true;
		}
	}
	return dp[n]=false;
}

bool Bellman_Ford()
{
	for (int i = 0; i < nodenum; ++i)  
		dis[i] = (i == 0 ? 0 : MAX);
	for (int i = 0; i < nodenum - 1; ++i)
	for (int j = 0; j < edgenum; ++j)
	if (dis[edge[j].v] > dis[edge[j].u] + edge[j].cost)  
	{
		dis[edge[j].v] = dis[edge[j].u] + edge[j].cost;
	}
	bool flag = 1;
	for (int i = 0; i < edgenum; ++i)
	if (dis[edge[i].v] > dis[edge[i].u] + edge[i].cost)
	{
		if (dfs(edge[i].u))
		{
			flag = 0;
			break;
		}
	}
	return flag;
}

int main()
{
	int casen;
	cin >> casen;
	for (int cas = 1; cas <= casen;cas++)
	{
		scanf("%d%d", &nodenum, &edgenum);
		memset(dis, 0, sizeof(dis));
		memset(vis, false, sizeof(vis));
		memset(dp, false, sizeof(dp));
		dp[0] = vis[0] = true;
		for (int i = 0; i < N; i++)
			G[i].clear();
		for (int i = 0; i < edgenum; i++)
		{
			scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].cost);
			G[edge[i].u].push_back(edge[i].v);
		}
		int ok=Bellman_Ford();

		if(!ok)printf("Case #%d: possible\n", cas);
		else printf("Case #%d: not possible\n", cas);
	}
}

UVALive - 6800 The Mountain of Gold?（Bellman-ford找负权回路，dfs）

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原文地址：http://blog.csdn.net/qq_18738333/article/details/45236231