poj 2253 Frogger

这题要说是最短路吧，其实更像DP，因为不需要把路径相加，而是更新为最小值。做这题最大的收获就是我才知道原来优先队列也能间接排序，自己写个cmp就行了，而且原理和sort的cmp一样。真的要感谢che学长。

#include <stdio.h>
#include <string.h>
#include <cmath>
#include <queue>
#include <vector>
using namespace std;
int n,m=1;
double d[205];
bool used[205];
typedef struct
{
    int x,y;
}Point;
Point a[205];
struct cmp
{
    bool operator()(int a , int b)
    {
        return d[a] > d[b] ;
    }
};
double dis(Point d,Point b)
{
    return sqrt(double(d.x-b.x)*(d.x-b.x)+(d.y-b.y)*(d.y-b.y));
}
double max(double a,double b)
{
    return a > b ?a :b ;
}
double Djstl()
{
    int t,i;
    priority_queue <int, vector<int>,cmp> q;
    memset(used,0,sizeof(used));
    for(i = 1; i <= n; i++)
        d[i] = 1000000;
    d[1] = 0;
    q.push(1);
    while(!q.empty())
    {
        t = q.top(); q.pop();
        if(used[t]) continue;
        used[t] = 1;
        for(i = 1; i <= n; i++)
        if(!used[i] && max(d[t],dis(a[t],a[i])) < d[i])
        {
            d[i] = max(d[t],dis(a[t],a[i]));
            q.push(i);
        }
    }
    return d[n];
}
int main()
{
    int i;
    double ans;
    while(scanf("%d",&n),n)
    {
        scanf("%d%d%d%d",&a[1].x,&a[1].y,&a[n].x,&a[n].y);
        for(i = 2; i < n; i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        ans = Djstl();
        printf("Scenario #%d\nFrog Distance = %.3lf\n\n",m++,ans);
    }
    return 0;
}