POJ-3259

Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 39402 Accepted: 14473

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

裸SPFA

#include <cstdio>#include <cstring>#include <iostream>#define Inf 0x7fffffffusing namespace std;int tot,x,y,t,n,m,w,f,z[800000],a[501],val[501],F[501];bool jud[501];struct thing{    int n,pre,v;} b[800000];void insert(int x,int y,int t){    tot++;    b[tot].n=y;    b[tot].pre=a[x];    b[tot].v=t;    a[x]=tot;}int main(){    cin>>f;    for(int T=1;T <= f;T++)    {        memset(F,0,sizeof(F));        memset(jud,0,sizeof(jud));        memset(a,0,sizeof(a));        tot=0;        cin>>n>>m>>w;        for(int i=1;i <= n;i++)            val[i]=Inf;        for(int i=1;i <= m;i++)        {            cin>>x>>y>>t;            insert(x,y,t);            insert(y,x,t);        }        for(int i=1;i <= w;i++)        {            cin>>x>>y>>t;            insert(x,y,-t);        }        for(int i=1;i <= n;i++)        {            insert(0,i,0);        }        int s=1,t=2;        z[s]=0;        jud[0]=true;        val[0]=0;        while(s != t)        {            int j=a[z[s]];            while (j)            {                if(val[b[j].n] > val[z[s]]+b[j].v)                {                    val[b[j].n]=val[z[s]]+b[j].v;                    if(!jud[b[j].n])                    {                        jud[b[j].n]=true;                        z[t]=b[j].n;                        F[b[j].n]++;                        if (F[b[j].n] >= n) break;                        t++;                    }                }                j=b[j].pre;            }            if (F[b[j].n] >= n)            {                val[0]=-1;                break;            }            jud[z[s]]=false;            s++;        }        if(val[0] < 0)        {            cout<<"YES"<<endl;        }        else cout<<"NO"<<endl;    }}