Wormholes POJ

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 53484 Accepted: 19924

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

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Link: http://poj.org/problem?id=3259

题意：给出多个farms，及farm之间的路径长度（双向），利用虫洞进行时光旅行（即路径为负），问是否存在负环。

Bell_Ford算法：

bool Bell_Ford(int s){  //s为起始地点    memset(dis,INF,sizeof(dis));    bool flag;    dis[s]=0;    for(int j=1;j<=nodeNum-1;j++)    {        bool flag=false;        for(int i=1;i<=edgeNum;i++){            if(dis[e[i].from]+e[i].cost<dis[e[i].to]){                dis[e[i].to]=dis[e[i].from]+e[i].cost;                flag=true;            }        }        if(!flag) break;    }

判断是否存在负环，存在返回真：

for(int i=1;i<=edgeNum;i++)        if(dis[e[i].from]+e[i].cost<dis[e[i].to])            return true;    return false;

ac代码：

#include<cstdio>#include<cstring>using namespace std;const int INF=0x3f3f3f3f;const int maxn=10000;int nodeNum,edgeNum;struct edge{    int from,to,cost;}e[maxn];int dis[maxn];bool Bell_Ford(){    memset(dis,INF,sizeof(dis));    bool flag;    dis[1]=0;    for(int j=1;j<=nodeNum-1;j++)    {        bool flag=false;        for(int i=1;i<=edgeNum;i++){            if(dis[e[i].from]+e[i].cost<dis[e[i].to]){                dis[e[i].to]=dis[e[i].from]+e[i].cost;                flag=true;            }        }        if(!flag) break;    }    //for(int i=1;i<=nodeNum;i++) printf("%d\n",dis[i]);    for(int i=1;i<=edgeNum;i++)        if(dis[e[i].from]+e[i].cost<dis[e[i].to])            return true;    return false;}int main(){    int t;    scanf("%d",&t);    while(t--){        int n,m,w;        scanf("%d%d%d",&n,&m,&w);        nodeNum=n;        edgeNum=m*2+w;        int f1,f2,cost;        int cnt=1;        for(int i=0;i<m;i++){            scanf("%d%d%d",&f1,&f2,&cost);            e[cnt].from=f1;            e[cnt].to=f2;            e[cnt].cost=cost;            cnt++;            e[cnt].from=f2;            e[cnt].to=f1;            e[cnt].cost=cost;            cnt++;        }        for(int i=0;i<w;i++){            scanf("%d%d%d",&f1,&f2,&cost);            //cost=cost*(-1);            e[cnt].from=f1;            e[cnt].to=f2;            e[cnt].cost=(-cost);            cnt++;        }        if(Bell_Ford()) printf("YES\n");        else printf("NO\n");    }}