POJ 2253 Frogger

这道题是要求石头1到石头2的最长跳跃距离的最小值，用了堆优化的dij来做，想不到好

久没写最短路，这玩意都写的不顺了。也可以用floyd求出任意两个石头间的距离，也就是

跳跃距离，然后找到1到2的最长路径的最小值。

/*Accepted    636K    0MS    C++    1453B    2012-07-26 11:51:06*/
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<cmath>
#include<iostream>
using namespace std;

typedef pair< double,int > pii;
const int MAXN = 1 << 8;
double x[MAXN], y[MAXN] ;
double g[MAXN][MAXN] ;
int n;

double dis( double a, double b)
{
    return sqrt(a * a + b * b);
}

void ReadGragh()
{
    for( int i = 1; i <= n; i ++)
        scanf( "%lf%lf", &x[i], &y[i]);
    for( int i = 1; i <= n; i ++)
        for( int j = i; j <= n; j ++)
        {
            g[i][j] = g[j][i] = dis( x[i] - x[j], y[i] - y[j]);
        }
}

double Dijkstra()
{
    priority_queue< pii, vector<pii>, greater<pii> > q;
    double dist[MAXN] = {0.0};
    for( int i = 2; i <= n; i ++)
        dist[i] = 1e120;
    q.push( make_pair(0.0, 1));
    while( !q.empty())
    {
        pii u = q.top(); q.pop();
        int x = u.second;
        if( x == 2) return dist[2];
        if( dist[x] != u.first) continue; //防止结点重复扩展
        for( int i = 2; i <= n; i ++)
        {
            if( i != x && dist[i] > max(dist[x], g[x][i]) )
            {
                dist[i] = max(dist[x], g[x][i]);
                q.push( make_pair(dist[i], i));
            }
        }
    }
    return 0;
}

int main()
{
    int cas = 1;
    while( scanf("%d", &n), n)
    {
        ReadGragh();
        printf( "Scenario #%d\n", cas ++);
        printf( "Frog Distance = %.3f\n\n", Dijkstra());
    }
}