POJ-3259 Wormholes（负权回路[Bellman-Ford]）

Wormholes

http://poj.org/problem?id=3259
Time Limit: 2000MS Memory Limit: 65536K   

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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.

题目大意：判断是否存在负权回路？

只以1号点为源点做Bellman-Ford即可判断图中是否存在负权回路（该负权回路不一定以1号点为起点）

例如：

1

3 0 2

2 3 1

3 2 1

这组数据会输出YES

调试时能发现，dis[2],dis[3]每次都在更新，但是是在INF的基础上更新，所以要判断1号点是否可打2，3时，要判断dis[2],dis[3]是否 大于 图中的最大权值和才行

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int MAXN=505;const int INF=0x3f3f3f3f;struct Edge {    int s,e,v;    Edge(int ss=0,int ee=0,int vv=0):s(ss),e(ee),v(vv) {}}u;int n,m,w,s,e,v;vector<Edge> edge;int dis[MAXN];bool Bellman_Ford(int sta) {//可判断负权回路    bool relaxed;    memset(dis,0x3f,sizeof(dis));    dis[sta]=0;    for(int i=1;i<n;++i) {        relaxed=false;        for(int j=0;j<edge.size();++j) {            if(dis[edge[j].s]+edge[j].v<dis[edge[j].e]) {//松弛                dis[edge[j].e]=dis[edge[j].s]+edge[j].v;                relaxed=true;            }        }        if(!relaxed) {//如果未更新，则不会再更新，且无负权回路            return false;        }    }    for(int j=0;j<edge.size();++j) {        if(dis[edge[j].s]+edge[j].v<dis[edge[j].e]) {//如果可以继续松弛，则存在负权回路            return true;        }    }    return false;}int main() {    int F;    scanf("%d",&F);    while(F-->0) {        edge.clear();        scanf("%d%d%d",&n,&m,&w);        for(int i=1;i<=m;++i) {            scanf("%d%d%d",&s,&e,&v);            edge.push_back(Edge(s,e,v));            edge.push_back(Edge(e,s,v));        }        for(int i=1;i<=w;++i) {            scanf("%d%d%d",&s,&e,&v);            edge.push_back(Edge(s,e,-v));        }        printf("%s\n",Bellman_Ford(1)?"YES":"NO");    }    return 0;}