acm-hdu1596解题报告

题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=1596

题目大意：类似图求最短路

解题思想：Floyd

代码：

#include <stdio.h>

#include <string.h>

#include <algorithm>

#include <iostream>

using namespace std;

const int inf = 100000000;

double map[1005][1005];

int max(int x,int y)

{

    return x>y?x:y;

}

void Floyd(int n)

{

    for(int k=0;k<n;k++)

    {

        for(int i=0;i<n;i++)

        {

            if(map[i][k]>0.00000001)

            for(int j=0;j<n;j++)

            {

                map[i][j]=max(map[i][k]*map[k][j],map[i][j]);

            }

        }

    }

}

int main()

{

    int n,m,i,j,x,y,z,start,end;

    while(scanf("%d",&n)!=EOF)

    {

        for(i=0;i<n;i++)

        {

            for(j=0;j<n;j++)

                 scanf("%lf",&map[i][j]);

        }

        Floyd(n);

        /* for(i=0;i<n;i++)

        {

            for(j=0;j<n;j++)

                cout<<map[i][j]<<" ";

            cout<<endl;

        }*/

        int m;cin>>m;

        for(int qw=0;qw<m;qw++)

        {

            scanf("%d%d",&start,&end);

            if(map[start-1][end-1]<0.00000001) printf("What a pity!\n");

            else printf("%.3lf\n",map[start-1][end-1]);

        }

    }

    return 0;

}