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 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
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 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.
本质是判断是否存在负环,直接用bellman_ford,SPFA都行。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
#define N 505
struct Edge{
int u,v,t;
Edge(int uu,int vv,int dd):u(uu),v(vv),t(dd){}
};
vector<Edge> edges;
int n,m,w;
int d[N];
#define INF 0x3f3f3f3f
bool bellman_ford(){
for (int i=1;i<=n;i++)
d[i]=INF;
d[1]=0;
for (int i=1;i<n;i++){
bool flag=false;
for (int j=0;j<edges.size();j++){
int u,v,t;
u=edges[j].u;
v=edges[j].v;
t=edges[j].t;
if (d[u]<INF && d[u]+t<d[v]){
d[v]=d[u]+t;
flag=true;
}
}
if (!flag)
return false;
}
for (int j=0;j<edges.size();j++){
int u,v,t;
u=edges[j].u;
v=edges[j].v;
t=edges[j].t;
if (d[u]<INF && d[u]+t<d[v]){
d[v]=d[u]+t;
return true;
}
}
return false;
}
int main(){
int T;
scanf("%d",&T);
while (T--){
edges.clear();
int u,v,d;
scanf("%d%d%d",&n,&m,&w);
for (int i=0;i<m;i++){
scanf("%d%d%d",&u,&v,&d);
edges.push_back(Edge(u,v,d));
edges.push_back(Edge(v,u,d));
}
for (int i=0;i<w;i++){
scanf("%d%d%d",&u,&v,&d);
edges.push_back(Edge(u,v,-d));
}
if (bellman_ford()){
printf("YES\n");
}else{
printf("NO\n");
}
}
return 0;
}