UVA - 104 Arbitrage

来源：互联网发布：公务员自动听课软件编辑：程序博客网时间：2024/04/29 18:28

题意：求给出n种国家的货币汇率，一定金额的某种货币经过一系列汇率变换后再换成原来货币，金额增加了，求出这样的一个变换，也就是最后的汇率大于等于1.01，要求变换步数最少。最多是转换更新n-1次，然后每转换一次就计算一次最短路，如果有满足题意的情况打印出路径

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 30;int path[MAXN][MAXN][MAXN],n;double dp[MAXN][MAXN][MAXN];void print(int i,int j,int p){    if (p == 0)        printf("%d",i);    else {        print(i,path[i][j][p],p-1);        printf(" %d",j);    }}void floyd(){    for (int p = 1; p < n; p++){        for (int k = 1; k <= n; k++)            for (int i = 1; i <= n; i++)                for (int j = 1; j <= n; j++){                    if (dp[i][k][p]*dp[k][j][1] > dp[i][j][p+1]+1e-12){                        dp[i][j][p+1] = dp[i][k][p]*dp[k][j][1];                        path[i][j][p+1] = k;                    }                }        for (int i = 1; i <= n; i++)            if (dp[i][i][p+1] > 1.01){                print(i,i,p+1);                printf("\n");                return;            }    }    printf("no arbitrage sequence exists\n");}int main(){    while (scanf("%d",&n) != EOF){        memset(dp,0,sizeof(dp));        for (int i = 1; i <= n; i++)            for (int j = 1; j <= n; j++)                if (i != j){                    scanf("%lf",&dp[i][j][1]);                    path[i][j][1] = i;                }        floyd();    }    return 0;}