uva 104(dp)

来源：互联网发布：在c 中执行sql语句编辑：程序博客网时间：2024/05/21 09:44

题意：给出了n种货币，然后给出了每种货币和其他n-1种货币的汇率，问是否能有一种货币换算情况使一种货币能从本身1经过换算后增加0.01以上，如果有输出最短的换算方式，否则输出no。

题解：dp题，用f[t][i][j]储存在第t次从货币i换算到货币j的最大换算出的钱，状态转移方程就是f[t][i][j] = max{f[t][i][j], f[t - 1][i][k] * m[k][j] },其中k是枚举的中转点，如果f[t][i][i] > 1.01说明成功，然后用path[t][i][j]存储第t次换算从货币i到j的中转点，递归打印路径。

#include <stdio.h>#include <string.h>const int N = 25;int n;double m[N][N];double f[N][N][N];int path[N][N][N];void print(int t, int i, int j) {if (t == 1) {printf("%d", i);return;}print(t - 1, i, path[t][i][j]);printf(" %d", path[t][i][j]);}bool dp() {for (int t = 2; t <= n; t++) {for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {double maxx = -1;int index;for (int k = 1; k <= n; k++)if (maxx < f[t - 1][i][k] * m[k][j]) {maxx = f[t - 1][i][k] * m[k][j];index = k;}f[t][i][j] = maxx;path[t][i][j] = index;}if (f[t][i][i] - 1.01 > 1e-9) {print(t, i, i);printf(" %d\n", i);return true;}}}return false;}int main() {while (scanf("%d", &n) != EOF) {for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (i != j) {scanf("%lf", &m[i][j]);f[1][i][j] = m[i][j];}int res = dp();if (!res)printf("no arbitrage sequence exists\n");}return 0;}

0 0