| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 34302 | Accepted: 12520 |
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 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
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 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.
#include <iostream>
#include <cstdio>
using namespace std; const int INF = 0xfffffff;
const int SIZE = ;
int D[];
int N,M,W,F;
struct Node
{
int from,to,cost;
}G[SIZE]; bool Bellman_Ford(int);
bool relax(int,int,int);
int main(void)
{
scanf("%d",&F);
while(F --)
{
scanf("%d%d%d",&N,&M,&W);
int i = ;
while(i < * M)
{
scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);
i ++;
G[i] = G[i - ];
swap(G[i].from,G[i].to);
i ++;
}
while(i < W + M * )
{
scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);
G[i].cost = -G[i].cost;
i ++;
}
bool flag = true;
for(int i = ;i <= N;i ++)
if(!Bellman_Ford(i))
{
puts("YES");
flag = false;
break;
}
if(flag)
puts("NO");
} return ;
} bool Bellman_Ford(int s)
{
fill(D,D + ,INF);
D[s] = ;
bool update; for(int i = ;i < N - ;i ++)
{
update = false;
for(int i = ;i < M * + W;i ++)
if(relax(G[i].from,G[i].to,G[i].cost))
update = true;
if(!update)
break;
}
for(int i = ;i < M * + W;i ++)
if(relax(G[i].from,G[i].to,G[i].cost))
return false;
return true;
} bool relax(int from,int to,int cost)
{
if(D[to] > D[from] + cost)
{
D[to] = D[from] + cost;
return true;
}
return false;
}
Bellman_Ford