poj3259 Wormholes【最短路-bellman-负环】

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 backT 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

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

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：走一条路会花费时间 走虫洞会时间倒流 问能不能走回起点的时候时间倒流

思路：其实就是判断有没有环 有环的话

看题的时候看了半天不知道起点到底是哪一个点

感觉现在bellman写的还挺顺手的了

一个加边的addedge函数 一个判断松弛的relax函数 

然后外面一个for循环遍历n-1次 里面的for循环松弛每一条边

再对每一条边判断能不能松弛 能就说明有环

有环其实就说明这个路径上有负的 

不然干吗要不停的重复走？？？

只有可以不断减小才会形成环

WA了一发是因为加边的时候的cnt没有初始化

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<cstring>
#include<queue>
#include<stack>
#define inf 0x3f3f3f3f

using namespace std;

int f, n, m, w, cnt;
struct edge{
    int s, e, t;
}path[5205];
long long d[505];

void addedge(int s, int e, int t)
{
    path[cnt].s = s;
    path[cnt].e = e;
    path[cnt].t = t;
    cnt++;
}

bool relax(int j)
{
    if(d[path[j].e] > d[path[j].s] + path[j].t){
        d[path[j].e] = d[path[j].s] + path[j].t;
        return true;
    }
    return false;
}

bool bellman(int sec)
{
    memset(d, inf, sizeof(d));
    d[sec] = 0;

    for(int i = 0; i < n - 1; i++){
        bool flag = false;
        for(int j = 0; j < cnt; j++){
            if(relax(j)) flag = true;
        }
        if(!flag) return false;
    }

    for(int i = 0; i < cnt; i++){
        if(relax(i)) return true;
    }
    return false;
}

int main()
{
    cin>>f;
    while(f--){
        cin>>n>>m>>w;
        cnt = 0;
        for(int i = 0; i < m; i++){
            int a, b, c;
            cin>>a>>b>>c;
            addedge(a, b, c);
            addedge(b, a, c);
        }
        for(int i = 0; i < w; i++){
            int a, b, c;
            cin>>a>>b>>c;
            addedge(a, b, -c);
        }

        if(bellman(1))
            cout<<"YES
";
        else
            cout<<"NO
";
    }
    return 0;
}