[USACO 2006 Dec G]Wormholes

NC25101. [USACO 2006 Dec G]Wormholes

描述

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 comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (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 to F (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.

输入描述

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

输出描述

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

示例1

输入:

输出:

NO
YES

说明:

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

原站题解

上次编辑到这里，代码来自缓存点击恢复默认模板

C++(clang++11) 解法, 执行用时: 493ms, 内存消耗: 2376K, 提交时间: 2021-02-08 19:40:57

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll inf=1e17;
ll a[510][510];
int n,m,w;
int main()
{
	int q;
	scanf("%d",&q);
	while(q--)
	{
		scanf("%d%d%d",&n,&m,&w);
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				a[i][j]=inf;
			}
			a[i][i]=0;
		}
		for(int i=0;i<m;i++)
		{
			int x,y;
			ll z;
			scanf("%d%d%lld",&x,&y,&z);
			a[x][y]=min(a[x][y],z);
			a[y][x]=min(a[y][x],z);
		}
		for(int i=0;i<w;i++)
		{
			int x,y;
			ll z;
			scanf("%d%d%lld",&x,&y,&z);
			z=-z;
			a[x][y]=min(a[x][y],z);
		}
		for(int k=1;k<=n;k++)
		{
			for(int i=1;i<=n;i++)
			{
				for(int j=1;j<=n;j++)
				{
					a[i][j]=min(a[i][j],a[i][k]+a[k][j]);
				}
			}
		}
		int k=0;
		for(int i=1;i<=n;i++)
		{
			if(a[i][i]<0)
			{
				k=1;
				break;
			}
		}
		if(k)
		printf("YES\n");
		else
		printf("NO\n");
	}
	return 0;
}

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