最短路(Bellman_Ford) POJ 1860 Currency Exchange

 

题目传送门

 1 /*  2  最短路(Bellman_Ford)：求负环的思路，但是反过来用，即找正环  3  详细解释：http://blog.csdn.net/lyy289065406/article/details/6645778  4 */  5 #include <cstdio>  6 #include <iostream>  7 #include <algorithm>  8 #include <cstring>  9 #include <vector> 10 #include <cmath> 11 #include <queue> 12 #include <map> 13 #include <set> 14 using namespace std; 15 16 const int MAXN = 100 + 10; 17 const int INF = 0x3f3f3f3f; 18 const double EPS = 1e-8; 19 struct NODE 20 { 21 int u, v; 22 double r, c; 23 }node[MAXN*2]; 24 double d[MAXN]; 25 26 bool Bellman_Ford(int s, int n, int tot, double V) 27 { 28 memset (d, 0, sizeof (d)); 29 d[s] = V; 30 bool flag; 31 for (int i=1; i<=n-1; ++i) 32  { 33 flag = false; 34 for (int j=1; j<=tot; ++j) 35  { 36 NODE e = node[j]; 37 if (d[e.v] < (d[e.u] - e.c) * e.r) 38  { 39 d[e.v] = (d[e.u] - e.c) * e.r; flag = true; 40  } 41  } 42 if (!flag) break; 43  } 44 45 for (int i=1; i<=tot; ++i) 46  { 47 NODE e = node[i]; 48 if (d[e.v] < (d[e.u] - e.c) * e.r) return true; 49  } 50 51 return false; 52 } 53 54 int main(void) //POJ 1860 Currency Exchange 55 { 56 //freopen ("A.in", "r", stdin); 57 58 int n, m, s; 59 double V; 60 while (~scanf ("%d%d%d%lf", &n, &m, &s, &V)) 61  { 62 int tot = 0; 63 for (int i=1; i<=m; ++i) 64  { 65 int x, y; 66 double rab, cab, rba, cba; 67 scanf ("%d%d%lf%lf%lf%lf", &x, &y, &rab, &cab, &rba, &cba); 68 //printf ("%d %d %lf %lf %lf %lf\n", x, y, rab, cab, rba, cba); 69 node[++tot].u = x; node[tot].v = y; 70 node[tot].r = rab; node[tot].c = cab; 71 node[++tot].u = y; node[tot].v = x; 72 node[tot].r = rba; node[tot].c = cba; 73  } 74 75 if (Bellman_Ford (s, n, tot, V)) puts ("YES"); 76 else puts ("NO"); 77  } 78 79 return 0; 80 }