poj 3259 Wormholes（spfa）

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 50350 Accepted: 18590

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.

tips：spfa判是否含有负环，没优化，时间复杂度O(MN)

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<queue>using namespace std;int dis[555];struct edge{int u,v,w;};vector<edge>edges;vector<int>g[555];int n,m,w,book[555],cnt[555];void insert(int x,int y,int z){edges.push_back(edge{x,y,z});g[x].push_back(edges.size()-1);}int spfa(){memset(dis,0x3f,sizeof(dis));memset(book,0,sizeof(dis));memset(cnt,0,sizeof(cnt));queue<int>q;q.push(1);book[1]=1;dis[1]=0;while(!q.empty()){int x=q.front();q.pop();book[x]=0;for(int i=0;i<g[x].size();i++){edge e=edges[g[x][i]];if(dis[e.v]>dis[x]+e.w){dis[e.v]=dis[x]+e.w;if(!book[e.v]){book[e.v]=1;q.push(e.v);if(++cnt[e.v]>n)return 1;}}}}return 0;}int main(){int t;cin>>t;while(t--){cin>>n>>m>>w;edges.clear();for(int i=0;i<=n;i++)g[i].clear();for(int i=1;i<=m;i++){int x,y,z;cin>>x>>y>>z;insert(x,y,z);insert(y,x,z);}for(int i=1;i<=w;i++){int x,y,z;cin>>x>>y>>z;insert(x,y,-z);}if(spfa())cout<<"YES"<<endl;else cout<<"NO"<<endl; }return 0; }