POJ3259 Wormholes(BF)

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 31457 Accepted: 11448

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.

Source

USACO 2006 December Gold

题目大体意思    虫洞问题，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

就是求是否有负权环.

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<iostream>using namespace std;#define INF 0x3f3f3f3fstruct node{    int x,y,z;}q[100010];int n,m,k;int t;int num[100001];void add(int x,int y,int z){    q[t].x = x;    q[t].y = y;    q[t++].z = z;    q[t].x = y;    q[t].y = x;    q[t++].z = z;}int BF(){    for(int i=0;i<=n+1;i++)    {        num[i] = INF;    }    num[0]=0;    int flag = 0;    for(int i=0;i<n-1;i++)    {        flag = 0;        for(int j=0;j<t;j++)        {            if(num[q[j].y]>num[q[j].x] + q[j].z)            {                num[q[j].y] = num[q[j].x] + q[j].z;                flag = 1;            }        }        if(flag == 0)        {            break;        }    }    for(int i=0;i<t;i++)    {        if(num[q[i].y]>num[q[i].x] + q[i].z)        {            return 1;        }    }    return 0;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        t = 0;        scanf("%d%d%d",&n,&m,&k);        for(int i=0;i<m;i++)        {            int x, y,z;            scanf("%d%d%d",&x,&y,&z);            add(x,y,z);        }        for(int i=0;i<k;i++)        {            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            q[t].x = x;            q[t].y = y;            q[t++].z = -z;        }        if(!BF())        {            cout << "NO" << endl;        }        else        {            cout << "YES" << endl;        }    }    return 0;}