POJ 3259 Wormholes Bellman题解

本题就是需要检查有没有负环存在于路径中，使用Bellman Ford算法可以检查是否有负环存在。

算法很简单，就是在Bellman Ford后面增加一个循环判断就可以了。

题目故事很奇怪，小心读题。

#include 
#include 
#include 

const int MAX_N = 501;
const int MAX_M = 2501;
const int MAX_W = 201;

struct Edge
{
	int src, des, wei;
	//Edge(int s, int d, int w) : src(s), des(d), wei(w) {}
};

Edge edge[(MAX_M<<1)+MAX_W];
int dist[MAX_N];

int N, M, W, F;

bool cycleBellmanFord()
{
	for (int i = 1; i <= N; i++) dist[i] = INT_MAX;
	dist[1] = 0;
	for (int i = 1; i < N; i++)
	{
		bool seperate = true;
		for (int j = 0; j < (M<<1)+W; j++)
		{			
			if (dist[edge[j].src] != INT_MAX && 
				dist[edge[j].src]+edge[j].wei < dist[edge[j].des])
			{
				dist[edge[j].des] = dist[edge[j].src]+edge[j].wei;
				seperate = false;
			}			
		}
		if (seperate) break;
	}
	for (int j = 0; j < (M<<1)+W; j++)
	{
		if ( dist[edge[j].src] != INT_MAX &&
			dist[edge[j].src]+edge[j].wei < dist[edge[j].des]) return true;
	}
	return false;
}

int main()
{
	scanf("%d", &F);
	while (F--)
	{
		scanf("%d %d %d", &N, &M, &W);
		int i = 0;
		for ( ; i < (M<<1); i++)
		{
			scanf("%d %d %d", &edge[i].src, &edge[i].des, &edge[i].wei);
			i++;
			edge[i].des = edge[i-1].src;
			edge[i].src = edge[i-1].des;
			edge[i].wei = edge[i-1].wei;
		}
		for ( ; i < (M<<1)+W; i++)
		{
			scanf("%d %d %d", &edge[i].src, &edge[i].des, &edge[i].wei);
			edge[i].wei = -edge[i].wei;
		}
		if (cycleBellmanFord()) puts("YES");
		else puts("NO");
	}
	return 0;
}