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

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;
}