POJ 2253 Frogger （floyed，路径最小值最大化）

POJ 2253 Frogger （floyed，路径最小值最大化）

vj链接

Solution

直接套 floyed 模板，更新的时候令 $dis[i][j]=max(dis[i][j],min(dis[i][k],dis[k][j]))$

代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<cmath>
#include<string>
#include<queue>
#include<set>
//#define int long long
using namespace std;
typedef pair<int, int> pii;
typedef double dd;
typedef long long ll;
const int MAXN = 210;
const int MAXM = 400010;
const dd eps = 1e-6;
const int mod = 100003;

int n;
pair<int, int> a[MAXN];
dd dis[MAXN][MAXN];

int main()
{
    int cnt = 0;
    while(cin >> n)
    {
        if(n==0)
            break;
        for (int i = 1; i <= n;i++)
        {
            cin >> a[i].first >> a[i].second;
        }
        for (int i = 1; i <= n;i++)
        {
            for (int j = i + 1; j <= n;j++)
            {
                //dis[i][j] = dis[j][i] = sqrt(pow(a[i].first - a[j].first, 2) + pow(a[i].second - a[j].second, 2));
                dis[i][j] = dis[j][i] = sqrt((dd)(a[i].first - a[j].first) * (dd)(a[i].first - a[j].first) + (a[i].second - a[j].second) * (a[i].second - a[j].second));
            }
        }
        for (int k = 1; k <= n;k++)
        {
            for (int i = 1; i <= n;i++)
            {
                for (int j = 1; j <= n;j++)
                    dis[i][j] = min(dis[i][j], max(dis[i][k], dis[k][j]));
            }
        }
        printf("Scenario #%d\n", ++cnt);
        printf("Frog Distance = %.3lf\n\n", dis[1][2]);
    }
}