POJ 3259 Wormholes (Bellman)

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23270		Accepted: 8301

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.

Source

USACO 2006 December Gold

 

 

题意：John的农场里field块地，path条路连接两块地，hole个虫洞，虫洞是一条单向路，不但会把你传送到目的地，而且时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。

思路：用bellman-ford 判断有没有负权回路，如果有他就能看到自己。 不过，我认为应该判断每个点有没有负权回路，而不仅仅只判断第一个点就行了（如果某位大牛路过看到，觉得理解不对 希望多多指教）

 

 http://www.cnblogs.com/Jason-Damon/archive/2012/04/21/2460850.html

 

#include<iostream>

#include<cstdio>

#include<cstring>



using namespace std;



const int VM=520;

const int EM=2520;

const int INF=0x3f3f3f3f;



struct Edge{

    int u,v;

    int cap;

}edge[EM<<1];



int n,m,k;

int cnt,dis[VM];



void addedge(int cu,int cv,int cw){

    edge[cnt].u=cu;     edge[cnt].v=cv;     edge[cnt].cap=cw;

    cnt++;

}



int Bellman_ford(){

    for(int i=1;i<=n;i++)

        dis[i]=INF;

    dis[1]=0;

    for(int i=1;i<n;i++)    //n-1次松弛

        for(int j=0;j<cnt;j++)  //枚举每条边

            if(dis[edge[j].v]>dis[edge[j].u]+edge[j].cap)

                dis[edge[j].v]=dis[edge[j].u]+edge[j].cap;

    for(int j=0;j<cnt;j++)

        if(dis[edge[j].v]>dis[edge[j].u]+edge[j].cap)   //判断是否存在负权边

            return 0;

    return 1;

}



int main(){



    //freopen("input.txt","r",stdin);



    int t;

    scanf("%d",&t);

    while(t--){

        scanf("%d%d%d",&n,&m,&k);

        cnt=0;

        int u,v,w;

        while(m--){

            scanf("%d%d%d",&u,&v,&w);

            addedge(u,v,w);

            addedge(v,u,w);

        }

        while(k--){

            scanf("%d%d%d",&u,&v,&w);

            addedge(u,v,-w);

        }

        int ans=Bellman_ford();

        printf("%s\n",ans==0?"YES":"NO");

    }

    return 0;

}

 

 下面这个代码稍快一点：

#include<iostream>

#include<cstdio>

#include<cstring>



using namespace std;



const int VM=520;

const int EM=2520;

const int INF=0x3f3f3f3f;



struct node{

    int u,v;

    int cap;

}edge[EM<<1];



int n,m,k;

int cnt,dis[VM];



void addedge(int cu,int cv,int cw){

    edge[cnt].u=cu;     edge[cnt].v=cv;     edge[cnt].cap=cw;

    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].cap){

                dis[edge[j].v]=dis[edge[j].u]+edge[j].cap;

                flag=1;

            }   

        if(!flag)    //优化

            break;

    }

    return i==n+1;  //相等则存在正环

}



int main(){



    //freopen("input.txt","r",stdin);



    int t;

    scanf("%d",&t);

    while(t--){

        scanf("%d%d%d",&n,&m,&k);

        cnt=0;

        int u,v,w;

        while(m--){

            scanf("%d%d%d",&u,&v,&w);

            addedge(u,v,w);

            addedge(v,u,w);

        }

        while(k--){

            scanf("%d%d%d",&u,&v,&w);

            addedge(u,v,-w);

        }

        int ans=Bellman_ford();

        printf("%s\n",ans==1?"YES":"NO");

    }

    return 0;

}