Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 32474 Accepted: 11808
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
/*Source Code
Problem: 3259 User: Grant Yuan
Memory: 224K Time: 125MS
Language: C++ Result: Accepted*/
Source Code
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m,w;
struct edges
{
int from,to,cost;
};
int d[507];
edges edge[6000];
int all=0;
bool Bellman_Ford()
{
bool flag;
memset(d,0,sizeof(d));
for(int i=1;i<=n;i++)
{
flag=false;
for(int k=0;k<all;k++)
{
if(d[edge[k].to]>d[edge[k].from]+edge[k].cost){
flag=true;
d[edge[k].to]=d[edge[k].from]+edge[k].cost;
}
}
if(i==n&&flag) return true;
}
return false;
}
int main()
{
// freopen("in.txt","r",stdin);
int f;int a,b,c;
scanf("%d",&f);
while(f--){
memset(edge,0,sizeof(edge));
all=0;
scanf("%d%d%d",&n,&m,&w);
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
edge[all].from=a;edge[all].to=b;edge[all++].cost=c;
edge[all].from=b;edge[all].to=a;edge[all++].cost=c;
}
for(int i=0;i<w;i++)
{
scanf("%d%d%d",&a,&b,&c);
edge[all].from=a;edge[all].to=b;edge[all++].cost=-c;
}
bool ans=Bellman_Ford();
if(ans) printf("YES
");
else printf("NO
");
}
return 0;
}