Poj 3259 Wormholes

1.Link：

http://poj.org/problem?id=3259

2.Content：

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 32078 Accepted: 11651

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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

3.Method：

Bellman算法的模板题目，主要用来找出负环是否存在，与dis的初始值没有什么关系。

4.Code：

 1 #include <iostream> 2 #include <cstring> 3  4 using namespace std; 5  6 //const int Max_d = 0; 7  8 struct Road 9 {10     int s;11     int e;12     int t;13 };14 15 int main()16 {17     //freopen("D://input.txt","r",stdin);18 19     int i,j;20 21     int ff;22     cin >> ff;23 24     int n,m,w;25     while(ff--)26     {27         cin >> n >> m >> w;28 29         Road *arr_road = new Road[m * 2 + w];30 31         int s,e,t;32         for(i = 0; i < m; ++i)33         {34             cin >> s >> e >> t;35             36             arr_road[i * 2].s = s;37             arr_road[i * 2].e = e;38             arr_road[i * 2].t = t;39 40             arr_road[i * 2 + 1].s = e;41             arr_road[i * 2 + 1].e = s;42             arr_road[i * 2 + 1].t = t;43         }44         for(i = 0; i < w; ++i)45         {46             cin >> s >> e >> t;47 48             arr_road[m * 2 + i].s = s;49             arr_road[m * 2 + i].e = e;50             arr_road[m * 2 + i].t = -t;51         }52 53         //for(i = 0; i < m * 2 + w; ++i)54         //{55         //    cout << i << " " << arr_road[i].s << " " << arr_road[i].e << " " << arr_road[i].t << endl;56         //}57         //cout << endl;58 59         int *arr_d = new int[n];60         //for(i = 0; i < n; ++i) arr_d[i] = Max_d;61         memset(arr_d,0,sizeof(int) * n);62 63         //Bellman-Ford64         bool flag;65         for(i = 0; i < n - 1; ++i)66         {67             flag = false;68             for(j = 0; j < 2 * m + w; ++j)69             {70                 if(arr_d[arr_road[j].e] > arr_d[arr_road[j].s] + arr_road[j].t)71                 {72                     arr_d[arr_road[j].e] = arr_d[arr_road[j].s] + arr_road[j].t;73                     flag = true;74                 }75             }76             if(!flag) break;77         }78 79         for(j = 0; j < 2 * m + w; ++j)80         {81             if(arr_d[arr_road[j].e] > arr_d[arr_road[j].s] + arr_road[j].t) break;82         }83 84         if(j < 2 * m + w) cout << "YES" << endl;85         else cout << "NO" << endl;86 87         delete [] arr_d;88         delete [] arr_road;89     }90 91     return 0;92 }

 

5.Reference：