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 0x7fffffff
using 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;
}
}