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

 

 

題意：John的農場里field塊地，path條路連接兩塊地，hole個蟲洞，蟲洞是一條單向路，不但會把你傳送到目的地，而且時間會倒退Ts。我們的任務是知道會不會在從某塊地出發后又回來，看到了離開之前的自己。

思路：用bellman-ford 判斷有沒有負權回路，如果有他就能看到自己。 不過，我認為應該判斷每個點有沒有負權回路，而不僅僅只判斷第一個點就行了（如果某位大牛路過看到，覺得理解不對 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++)  //枚舉每條邊
            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)   //判斷是否存在負權邊
            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;
}

 

 下面這個代碼稍快一點：

#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)    //優化
            break;
    }
    return i==n+1;  //相等則存在正環
}

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;
}