题目

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.

题意

有一个人，他喜欢时间旅行，现在有一些虫洞，可以回到过去，现在有 n 个点， m 条边，代表现在可以走的通路，比如从 a 到 b 和从 b 到 a 需要花费 c 时间，现在在地上出现了 w 个虫洞，虫洞的意义就是你从 a 到 b 话费的时间是 −c (时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前。现在让你判断他能不能回到从前。

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思路

题目的数据给出了每个点的坐标，这样就可以构成一张图，用 Bellman−Ford 算法判断该图中是否存在负环，如果存在，输出“YES”,否则输出“NO”。题目数据比较水，没有考虑重边的情况。。。可以套模板

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代码

#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<queue> #include<stack> #include<vector> #include<cmath> #include<set> #include<map> #include<cstdlib> #include<functional> #include<climits> #include<cctype> #include<iomanip> using namespace std; typedef long long ll; #define INF 0x3f3f3f3f #define mod 1e9+7 #define clr(a,x) memset(a,x,sizeof(a)) const double eps = 1e-6; const int MAX_N=10000; const int MAX_E=10000; int From[MAX_E],To[MAX_E],W[MAX_E]; int dis[MAX_N],tot; void init() { tot=0; } void add_edge(int u,int v,int d) { From[tot]=u; To[tot]=v; W[tot++]=d; } bool Bellman_ford(int s,int n) { clr(dis,0x3f); dis[s]=0; for(int k=0;k<n-1;k++) { bool relaxed=0; for(int i=0;i<tot;i++) { int x=From[i],y=To[i]; if(dis[y]>dis[x]+W[i]) { dis[y]=dis[x]+W[i]; relaxed=1; } } if(!relaxed) break; } for(int i=0;i<tot;i++) if(dis[To[i]]>dis[From[i]]+W[i]) return 1; return 0; } int main() { int f,n,m,t,u,v,w; cin>>f; while(f--) { int num=0; init(); cin>>n>>m>>t; //*************************************建图 for(int i=0;i<m;i++) { cin>>u>>v>>w; add_edge(u,v,w); add_edge(v,u,w); num+=2; //记录边的个数 } for(int i=0;i<t;i++) { cin>>u>>v>>w; add_edge(u,v,-w); } //*************************************建图 bool ans=Bellman_ford(1,num); if(ans) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0; }