学步园

    做本题的同学可以做 1162 Eddy's picture（hud）

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
#define MAX 5000

struct dis
{
    double x, y;
}N[101];
//     权值    路程总和
double c[MAX], sum;
//   并查集  排序           端点
int  p[105], r[MAX],  n, m, a[MAX], b[MAX], nc;

double cmp(const int x, const int  y)
{
    return c[x]<c[y];
}
int find(int n)
{
    return p[n]==n?n:find(p[n]);
}
void nKurskal()
{
    int i;
    for (i=1; i<=n; i++)
        p[i] = i;
    for (i=1; i<=m; i++)
        r[i] = i;
    sort(r+1, r+m+1, cmp);
    for (i=1; i<=m; i++)
    {
        int t = r[i];
        int x = find(a[t]);
        int y = find(b[t]);
        if (x!=y && c[t]>=10.0 && c[t]<=1000.0)//条件判断
        {
            sum += c[t];
            p[x] = y;
            nc++;  //计算是否包含所有的顶点
        }
    }
}

int main()
{
    int i, j, num;
    scanf("%d", &num);
    while (num--)
    {
        scanf("%d", &n);
        for (i=1; i<=n; i++)
            cin>>N[i].x>>N[i].y;
        nc = 1;
        for (i=1; i<=n; i++)
            for (j=i+1; j<=n; j++)
            {
                a[nc] = i;
                b[nc] = j;
                c[nc] = sqrt((N[i].x-N[j].x)*(N[i].x-N[j].x)+(N[i].y-N[j].y)*(N[i].y-N[j].y)); //计算距离 权值
                nc++;
            }
            m = n * (n-1) / 2;//村庄数量
            sum = 0;
            nc = 1;
            nKurskal();
            if (nc==n)
                printf("%.1f\n", sum*100.0);
            else
                printf("oh!\n");
    }
return 0;
}