poj 2253 Frogger

这道题我是用并查集处理的，将所有边排序，当生成树里出现第一个点，第二个点时（或第0点第1个点），结束并查集，输出已处理边里的最大值就可以了。

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define MAXN 210
int n,g,h,u[MAXN*MAXN],v[2*MAXN*MAXN],p[MAXN],r[2*MAXN*MAXN];
double x[MAXN],y[MAXN],w[2*MAXN*MAXN],flag[2*MAXN*MAXN];

int cmp(const void *_p, const void *_q)
{
    int *p = (int *)_p;
    int *q = (int *)_q;
    return w[*p]>w[*q] ? 1 : -1;
}
int find(int x2){return p[x2] == x2 ? x2 : p[x2] = find(p[x2]);}
void kruskal()
{
    double ans = -1;
    int ok1 = 0, ok2 = 0;
    for(int i = 0; i < n; i ++) p[i] = i;
    for(int i = 0; i < h; i ++) r[i] = i;
    qsort(r,h,sizeof(r[0]),cmp);
    g = 0;
    for(int i = 0; i < h; i ++)
    {
        int e = r[i];
        int x1 = find(u[e]), y1 = find(v[e]);
        if(find(0) == find(1))break;
        if(x1 != y1)
        {
            p[x1] = y1;
            flag[g ++] = w[e];
         }
    }
    for(int i = 0; i < g; i ++)
    {
        if(flag[i] > ans) ans = flag[i];
    }
    printf("Frog Distance = %.3f\n\n",ans);
}
void init()
{
    int num = 1;
    while(~scanf("%d",&n))
    {
        if(n == 0) break;
        for(int i = 0; i < n; i ++)
            scanf("%lf%lf",&x[i], &y[i]);
        h = 0;
        for(int i = 0; i < n; i ++)
            for(int j = i+1; j < n; j ++)
            {
                double dis = sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
                u[h] = i, v[h] = j, w[h++] = dis;
                u[h] = j; v[h] = i, w[h++] = dis;
            }
            printf("Scenario #%d\n",num++);
        kruskal();    
    }    
}

int main()
{
    init();
    return 0;
}