Wormholes
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 30169 |
|
Accepted: 10914 |
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
AC代码：
#include
using namespace std;
struct Point{
int s,e,t;
}a[10000];
int se;
int n,m,w;
int bell_man(int start){
int dis[10000];
for(int i=1;i<=n;i++)
dis[i]=999999;
dis[start]=0;
for(int i=1;i
dis[a[j].s] + a[j].t ? dis[a[j].s] + a[j].t : dis[a[j].e]; for(int i=0;i
dis[a[i].s] + a[i].t) return 1; } return 0; } int main(){ int T; cin>>T; while(T--){ se=0; cin>>n>>m>>w; for(int i=0;i
>s>>e>>t; a[se].s=s; a[se].e=e; a[se++].t=t; a[se].s=e; a[se].e=s; a[se++].t=t; } for(int i=0;i
>s>>e>>t; a[se].s=s; a[se].e=e; a[se++].t=-t; } //int k; //for(k=1;k<=n;k++){ //其实正确的起点应该要历遍所有点，但是这样的超时了 //这个题目只要1点就可以了，算是题目的一个很大
漏洞吧，数据太水了 if(bell_man(1)){ cout<<"YES"<
n) else cout<<"NO"<