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 backT 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.
题目意思; 有一个农场中有农田和虫洞,虫洞可以倒退一段时间,判断一个人能否回到出发前;
解题思路:判断是否存在负权回路;
1 #include<iostream> 2 #include <string.h> 3 #include <math.h> 4 #include <stdio.h> 5 using namespace std; 6 7 const int MAX = 10000 + 100; 8 long long MAX1 = 1e10; 9 long long dist[MAX]; 10 int add; 11 int n1,n2,n3; 12 13 struct S 14 { 15 int a,b; 16 int len; 17 }; 18 S Map[MAX]; 19 bool B() 20 { 21 for(int i = 1;i <= n1;i++) 22 dist[i] = MAX1; 23 dist[1] = 0; 24 25 for(int i =1;i <=n1;i++) 26 { 27 bool flag = false; 28 for(int i = 0;i < add;i++) 29 { 30 if( dist[ Map[i].b ] > dist[Map[i].a] + Map[i].len ) 31 { 32 dist[ Map[i].b ] = dist[Map[i].a] + Map[i].len; 33 flag = true; 34 } 35 36 } 37 if(!flag) 38 break; 39 } 40 41 for(int i = 0;i <add;i++) 42 if(dist[ Map[i].b ] > dist[Map[i].a] + Map[i].len) 43 return true; 44 return false; 45 46 } 47 48 49 int main() 50 { 51 int N; 52 cin>>N; 53 54 while(N--) 55 { 56 cin>>n1>>n2>>n3; 57 58 add = 0; 59 for(int i =1;i <= n2;i++) 60 { 61 int x,y,len; 62 cin>>x>>y>>len; 63 Map[add].a = x;Map[add].b=y;Map[add++].len = len; 64 Map[add].a = y;Map[add].b=x;Map[add++].len = len; 65 66 } 67 for(int i = 1;i <= n3;i++) 68 { 69 int x,y,len; 70 cin>>x>>y>>len; 71 Map[add].a = x;Map[add].b=y;Map[add++].len = -len; 72 } 73 74 if( B() ) 75 cout<<"YES"<<endl; 76 else 77 cout<<"NO"<<endl; 78 79 } 80 81 82 return 0; 83 }