最短路(Bellman_Ford) POJ 3259 Wormholes

 

题目传送门

 1 /*  2  题意：一张有双方向连通和单方向连通的图，单方向的是负权值，问是否能回到过去(权值和为负)  3  Bellman_Ford：循环n-1次松弛操作，再判断是否存在负权回路(因为如果有会一直减下去)  4  注意：双方向连通要把边起点终点互换后的边加上  5 */  6 #include <cstdio>  7 #include <iostream>  8 #include <algorithm>  9 #include <cmath> 10 #include <cstring> 11 #include <string> 12 #include <map> 13 #include <vector> 14 #include <queue> 15 using namespace std; 16 17 const int MAXN = 6000 + 10; 18 const int INF = 0x3f3f3f3f; 19 int d[MAXN]; 20 struct NODE 21 { 22 int from, to, cost; 23 }node[MAXN]; 24 25 bool Bellman_Ford(int s, int n, int m) 26 { 27 int x, y; 28 d[s] = 0; 29 for (int k=1; k<=n-1; ++k) 30  { 31 for (int i=1; i<=m; ++i) 32  { 33 NODE e = node[i]; 34 d[e.to] = min (d[e.to], d[e.from] + e.cost); 35  } 36  } 37 38 for (int i=1; i<=m; ++i) 39  { 40 NODE e = node[i]; 41 if (d[e.to] > d[e.from] + e.cost) return false; 42  } 43 44 return true; 45 } 46 47 void work(int n, int m) 48 { 49 bool flag = false; 50 flag = Bellman_Ford (1, n, m); 51 52 (!flag) ? puts ("YES") : puts ("NO"); 53 } 54 55 int main(void) //POJ 3259 Wormholes 56 { 57 //freopen ("B.in", "r", stdin); 58 59 int N, M, W; 60 int t; 61 scanf ("%d", &t); 62 while (t--) 63  { 64 scanf ("%d%d%d", &N, &M, &W); 65 for (int i=1; i<=N; ++i) d[i] = INF; 66 67 int x, y, z; 68 for (int i=1; i<=2*M; ++i) 69  { 70 scanf ("%d%d%d", &x, &y, &z); 71 node[i].from = x; node[i].to = y; node[i].cost = z; 72 node[++i].from = y; node[i].to = x; node[i].cost = z; 73  } 74 75 for (int i=2*M+1; i<=2*M+W; ++i) 76  { 77 scanf ("%d%d%d", &x, &y, &z); 78 node[i].from = x; node[i].to = y; node[i].cost = -z; 79  } 80 81 work (N, 2*M+W); 82  } 83 84 return 0; 85 }