poj 3259 Wormholes（判断图中是不是存在负环）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 59278		Accepted: 22160

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.

//题意:给你n个区域，有m条双向路把这n个区域连起来,其中还有w个单向的虫洞可以当路来用(这样就有2*m+w条路)；
//接下来的m行输入这m条路，每条路由起点s,终点e,话费时间t构成;
//再接下来的w行表示w个虫洞，每个虫洞由起点s,终点e，以及从s到e可以让你的时间回退多少（就是这条边的权值是负的,它是用来减时间的）；
//这样题目就来了,问你他走着走着是否可以回到起始时间之前，就是问你图中是否存在负环.
//这不就很简单了吗? 
#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cmath>
#define maxn1 505
#define maxn2 5400
#define INF 0x3f3f3f3f
using namespace std;
int n, m, w; //n个field，m条双向道路,w条单向虫洞;
struct edge{
    int from, to, cost;
}G[maxn2];
int dis[maxn1];
int len;
bool find_negative_loop(){
    memset(dis, 0, sizeof(dis));

    for(int i = 0; i < n; i++){
        for(int j = 0; j < len; j++){
            edge e = G[j];
            if(dis[e.to] > dis[e.from]+e.cost){
                dis[e.to] = dis[e.from]+e.cost;

                if(i == n-1)      //如果从第0到第n-1总共n次都更新了,说明存在负环(不存在负环的图，第一层for循环最多进行n-1次)
                    return true;
            }
        }
    }
    return false;
}
int main(){
    int s, e, v;
    int f;
    cin >> f;
    while(f--){
        scanf("%d%d%d", &n, &m, &w);
        len = 0;
        for(int i = 0; i < m; i++){   //道路是双向的
            scanf("%d%d%d", &s, &e, &v);
            G[len].from = s;
            G[len].to = e;
            G[len].cost = v;
            len++;
            G[len].from = e;
            G[len].to = s;
            G[len].cost = v;
            len++;
        }
        for(int j = 0; j < w; j++){  //虫洞是单向的
            scanf("%d%d%d", &s, &e, &v);
            G[len].from = s;
            G[len].to = e;
            G[len].cost = -v;
            len++;
        }
        if(find_negative_loop()){
            printf("YES\n");
        }
        else
            printf("NO\n");
    }
    return 0;
}

poj 3259 Wormholes（判断图中是不是存在负环）

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