poj_3259Wormholes

最新推荐文章于 2021-10-26 22:16:24 发布

n3verl4nd

最新推荐文章于 2021-10-26 22:16:24 发布

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本文探讨了通过特定路径和虫洞实现回到过去的可能性，利用贝尔曼-福特算法判断是否存在负权回路，进而验证时间旅行的可行性。

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Wormholes

Time Limit:2000MS		Memory Limit:65536K
Total Submissions:24055		Accepted:8573

Description

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprisesN(1 ≤N≤ 500) fields conveniently numbered 1..N,M(1 ≤M≤ 2500) paths, andW(1 ≤W≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps toF(1 ≤F≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer,F.Ffarm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively:N,M, andW
Lines 2..M+1 of each farm: Three space-separated numbers (S,E,T) that describe, respectively: a bidirectional path betweenSandEthat requiresTseconds to traverse. Two fields might be connected by more than one path.
LinesM+2..M+W+1 of each farm: Three space-separated numbers (S,E,T) that describe, respectively: A one way path fromStoEthat also moves the traveler backTseconds.

Output

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

Sample Output

NO
YES

判断有无负环。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<cmath>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<queue>
#include <stack>
using namespace std;
#pragma warning(disable : 4996)
const int INF = 999999;  
const int MAXN = 10005;

typedef struct Node
{
	int v;//起点
	int u;//终点
	int w; 
}Node;
Node edge[MAXN];
int dist[MAXN];     //此处要特别注意，bellman-ford算法中不要使用0x7fffffff,为此wa了n次 
int edgenum, n, m, w;

bool BellmanFord(int s)
{
	int i, j;
	bool flag = false;
	for(i = 1; i <= n; ++i)
	{
		dist[i] = INF;        //其余点的距离设置为无穷
	}
	dist[s] = 0;             //源点的距离设置为0
	for(i = 1; i < n; ++i)
	{
		flag = false;       //优化：如果某次迭代中没有任何一个d值改变，尽可以立刻退出迭代而不需要把所有的n-1次迭代都做完
		for(j = 0; j < edgenum; ++j)
		{
			if(dist[edge[j].v] > dist[edge[j].u] + edge[j].w)
			{
				flag = true;
				dist[edge[j].v] = dist[edge[j].u] + edge[j].w;
			}
		}
		if(!flag)
		{
			break;
		}
	}
	for(i = 0; i < edgenum; ++i)
	{
		if(dist[edge[i].v] > dist[edge[i].u] + edge[i].w)
		{
			return false;//存在负环
		}
	}
	return true;//不存在负环
}

int main()
{
	freopen("in.txt", "r", stdin);
	int t, x, y, z;
	scanf("%d", &t);
	while (t--)
	{
		edgenum = 0;
		scanf("%d %d %d", &n, &m, &w);
		for(int i = 1; i <= m; i++)
		{
			scanf("%d %d %d", &x, &y, &z);
			edge[edgenum].u = x;
			edge[edgenum].v = y;
			edge[edgenum++].w = z;
			edge[edgenum].u = y;
			edge[edgenum].v = x;
			edge[edgenum++].w = z;
		}
		for(int i = 1; i <= w; i++)
		{
			scanf("%d %d %d", &x, &y, &z);
			edge[edgenum].u = x;
			edge[edgenum].v = y;
			edge[edgenum++].w = -z;
		}
		if(BellmanFord(1))
		{
			printf("NO\n");
		}
		else
		{
			printf("YES\n");
		}
	}
	return 0;
}