Poj 3259 wormholes

Wormholes

Time limit:2000 ms		Memory limit:65536 K
Total submissions:30766		Accepted:11157

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

NOYES

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.

Source

Usaco 2006 December gold

Question: there is a two-way positive side and a single negative side (the time of the worm hole is reversed). Ask if you can go back.

Idea: Use the bellman_ford algorithm to determine whether a negative weight ring exists.

#include <cstdio>#include <iostream>#include <algorithm>#include <cmath>#include <cstring>#include <stdlib.h>using namespace std;const int INF=1000000009;struct Node{    int s,e,t;}mp[5205];int d[505];int n,m,w;bool bellman_ford(){    memset(d,INF,sizeof(d));    d[1]=0;    for(int i=1;i<n;i++){        bool flag=true;        for(int j=1;j<=2*m+w;j++){            int a=mp[j].s,b=mp[j].e,c=mp[j].t;            if(d[b]>d[a]+c){                d[b]=d[a]+c;                flag=false;            }        }        if(flag) false;    }    for(int i=1;i<=2*m+w;i++){        int a=mp[i].s,b=mp[i].e,c=mp[i].t;        if(d[b]>d[a]+c)            return false;    }    return true;}int main(){    int t;    scanf("%d",&t);    while(t--){        memset(mp,0,sizeof(mp));        scanf("%d%d%d",&n,&m,&w);        int k=0;        for(int i=1;i<=m;i++){            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            ++k;            mp[k].s=x,mp[k].e=y,mp[k].t=z;            ++k;            mp[k].e=x,mp[k].s=y,mp[k].t=z;        }        for(int i=m+1;i<=m+w;i++){            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            ++k;            mp[k].s=x,mp[k].e=y,mp[k].t=-z;        }        if(!bellman_ford()) printf("YES\n");        else printf("NO\n");    }    return 0;}

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