UVA - 104 Arbitrage

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;
}