HDU 1217 Arbitrage

题意：

某些币种之间可以通过一定比率进行兑换，

问是否存在一种币种  经过若干次currency exchang回到原币种时，能够得到盈利。

即求 st * v1 * v2 * ... * vn * st 的一个最大值。

(1<=n<=30) 数据量较小，用动态规划思想的Floyd能求的最大值。

 

00445234

2010-07-02 11:40:04

Accepted

1008

46 MS

300 KB

Visual C++

Slave_wc

 

#include <iostream> #include <cstring> #include <cstdio> #include <string> #include <cmath> #include <map> using namespace std; #define INF 0x6fffffff #define MAXN 33 double maze[MAXN][MAXN]; map<string, int> hash_str; double max(double a, double b) { return (a > b ? a : b); } void Floyd(int n) { for (int k = 1; k <= n; k++) { for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { maze[i][j] = max(maze[i][k] * maze[k][j], maze[i][j]); } } } double max = 0; for (int i = 1; i <= n; i++) { if (maze[i][i] > max) { max = maze[i][i]; } } if (max > 1) { puts("Yes"); } else { puts("No"); } } int main() { int n, m, k, i, x, y; int test = 1; double w; char str[150], a[150], b[150]; while (scanf("%d", &n) != EOF) { if (n == 0) { break; } memset(maze, 0, sizeof(maze)); k = 1; for (i = 1; i <= n; i++) { scanf("%s", str); if (!hash_str[str]) { hash_str[str] = k++; } } scanf("%d", &m); bool flag = false; for (i = 1; i <= m; i++) { scanf("%s %lf %s", a, &w, b); x = hash_str[a]; y = hash_str[b]; maze[x][y] = w; if (x == y && w > 1) { flag = true; } } printf("Case %d: ", test++); if (flag) { puts("Yes"); } else { Floyd(n); } hash_str.clear(); } return 0; }