POJ - 3259----Wormholes（Bellman-Ford）

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 Input23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8Sample OutputNOYESHintFor 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.、

题目大意：有一个穿越时光的农夫，他有N个农场，每个农场有M条双向路，有W个单向虫洞，每条路有三个属性（s，e，t）分别代表的是从s到达e花费t时间，每个虫洞也有三个属性（s，e，t）分别代表的是s到达e花费-t时间，这个农夫想要满足自己的穿越时光的梦想，他想在他的农场里游荡一圈以后回到原点并且看到原来的自己，如果能满足他的梦想就输出“YES”，否者“NO”。
思路：Bellman-Ford算法(判断负权回路)

#include<algorithm>#include<iostream>#include<stdlib.h>#include<cstring>#include<cstdio>#include<cstdio>#include<cmath>#include<queue>using namespace std;const int MAX = 6000const int INF = 0x3f3f3f3f;//边集 struct edge{    int start,end,weigth;}E[MAX];int dis[MAX];   //出发点到其余点的最短距离 int edge,dot;   //边集数，点集数 //单源最短路BF算法(判断负权回路) bool BF(int start){    for(int i=1 ; i<=dot ; i++){        dis[i] = (i==start)?0:INF;    }    //在对每条边进行1遍松弛的时候，生成了从s出发，层次至多为1的那些树枝。    //也就是说，找到了与s至多有1条边相联的那些顶点的最短路径；    //对每条边进行第2遍松弛的时候，生成了第2层次的树枝，    //就是说找到了经过2条边相连的那些顶点的最短路径……。    //因为最短路径最多只包含|v|-1 条边，所以，只需要循环|v|-1 次。    //循环次数为  起点到终点所经过的边数  最多为点数减一条边      for(int i=1 ; i<=dot-1 ; i++){        //枚举所有边集         bool flag = true;        for(int j=1 ; j<=edge ; j++){            //这条边可以缩短路的长度，更新             if(dis[E[j].start]+E[j].weigth<dis[E[j].end]){                dis[E[j].end] = dis[E[j].start]+E[j].weigth;                flag = false;            }        }        //没有更新长度，说明已更新完         if(flag){            break;        }    }    //判断负权回路     //再次加入一遍边集，若还能更新 说明存在负权回路     for(int j=1 ; j<=edge ; j++){        if(dis[E[j].start]+E[j].weigth<dis[E[j].end]){            return true;        }    }    return false;}int main(void){    int T;    cin>>T;    while(T--){        //无向边,有向边         int num1,num2;        cin>>dot>>num1>>num2;        //输入权值(路径长度)        edge = 1;         for(int i=1 ; i<=num1 ; i++){              int x,y,w;            cin>>x>>y>>w;            E[edge].start = x;            E[edge].end = y;            E[edge++].weigth = w;            E[edge].start = y;            E[edge].end = x;            E[edge++].weigth = w;        }        for(int i=1 ; i<=num2 ; i++){              int x,y,w;            cin>>x>>y>>w;            E[edge].start = x;            E[edge].end = y;            E[edge++].weigth = -w;        }        edge--;        if(BF(1)){            cout<<"YES"<<endl;        }else{            cout<<"NO"<<endl;        }    }     return 0;}