poj 3259 Wormholes

最新推荐文章于 2021-08-13 10:28:22 发布

kimnoic

最新推荐文章于 2021-08-13 10:28:22 发布

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本文深入探讨了AI音视频处理领域中的视频分割与语义识别技术，介绍了视频分割的基本概念、算法及其应用，并详细阐述了语义识别在智能视频分析中的重要作用，包括其在自动驾驶、AR增强现实等领域的实际应用案例。

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Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34529		Accepted: 12610

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 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.

Input

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.

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

Hint

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.

Source

USACO 2006 December Gold

给一个图，判断有没有负环。

bellman_ford一下，看能不能超n

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <stdlib.h>
#include <vector>
using namespace std;
#define INF 1<<30
struct line
{
	int b,e;
	int t;
}e[6666];
int minn[1000];
int T,n,m,w,s,en,t,cnt;
bool relax(int p)
{
	int sum=minn[e[p].b]+e[p].t;
	if(sum<minn[e[p].e])
	{
		minn[e[p].e]=sum;
		return true;
	}
	return false;
}
bool bellman()
{
	for(int i=1;i<=n;i++) minn[i]=INF;
	bool flag;
	for(int i=1;;i++)
	{
		if(i>n) return true;
		flag=false;
		for(int j=1;j<=cnt;j++)
		{
			if(relax(j)) flag=true;
		}
		if(!flag) return false;
	}
}
int main()
{
	cin>>T;
	while(T--)
	{
		cnt=0;
		cin>>n>>m>>w;
        for(int i=1;i<=m;i++)
		{
			cin>>s>>en>>t;
			cnt++;
			e[cnt].b=s;e[cnt].e=en;e[cnt].t=t;
			cnt++;
			e[cnt].b=en;e[cnt].e=s;e[cnt].t=t;

		}
		for(int i=1;i<=w;i++)
		{
			cin>>s>>en>>t;
			cnt++;
			e[cnt].b=s;e[cnt].e=en;e[cnt].t=-t;
		}
		if(bellman()) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
	return 0;
}