Wormholes (Shortest _ bellman_ford)

Wormholes

Time limit:2000 ms		Memory limit:65536 K
Total submissions:31762		Accepted:11561

Description

While processing his own 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 comprisesN(1 ≤N≤ 500) fields conveniently numbered 1 ..N,M(1 ≤M≤ 2500) paths, andW(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 mapsF(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. FFarm 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 SAnd EThat requires TSeconds 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 STo EThat also moves the traveler back TSeconds.

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

NO

Yes
Hint
For Farm 1, FJ cannot travel back in time. for Farm 2, FJ cocould travel back in time by the cycle 1-> 2-> 3-> 1, arriving back at his starting location 1 second before he leaves. he cocould start from anywhere on the cycle to accomplish this. john's farm has n places, m links are connected to two places, W wormhole holes, and the wormhole holes are a one-way road that will send you to the destination before you leave, that is, TS will be regressed in the past. Our task is to know if we will return again after starting from a certain place and see ourselves before leaving. Idea: Check whether there is a negative ring. If there is a negative ring, it will prove that yes can be returned, and no can be output if there is no negative ring. You can use bellmanford to determine the negative ring.
#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>using namespace std;#define inf 0x3f3f3f3fstruct node{    int u,v,w;} edge[6010];int dis[510];int cnt;int n,m,W;void add_edge(int u,int v,int w){    edge[cnt].u=u;    edge[cnt].v=v;    edge[cnt].w=w;    cnt++;}int bellman_ford(){    int i,j;    for(i=1; i<=n; i++)        dis[i]=inf;    dis[1]=0;    for(i=1; i<n; i++)    {        int flag=0;        for(j=0; j<cnt; j++)        {            if(dis[edge[j].v]>dis[edge[j].u]+edge[j].w)            {                dis[edge[j].v]=dis[edge[j].u]+edge[j].w;                flag=1;            }        }        if(!flag)            break;    }    for(i=0; i<cnt; i++)        if(dis[edge[i].v]>dis[edge[i].u]+edge[i].w)            return 1;    return 0;}int main(){    int T;    int u,v,w;    scanf("%d",&T);    while(T--)    {        cnt=0;        scanf("%d %d %d",&n,&m,&W);        while(m--)        {            scanf("%d %d %d",&u,&v,&w);            add_edge(u,v,w);            add_edge(v,u,w);        }        while(W--)        {            scanf("%d %d %d",&u,&v,&w);            add_edge(u,v,-w);        }        if(bellman_ford())            printf("YES\n");        else            printf("NO\n");    }    return 0;}

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Wormholes (Shortest _ bellman_ford)