​​传送门​​​ 思路：给你坐标点，让你求一条经过每个点的最短路径。
题意：最小生成树做即可。注意特判条件和位置。应放在kru中。

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
const int maxn = 1e6 + 5;
using namespace std;
struct node {
    double  u, v;
    double w;
} e[maxn];
int p[maxn];
bool cmp(node x, node y) {
    return x.w < y.w;
}
int find(int x) { ///并查集压缩路径//加边法
    ///return p[x]==x?x:find(p[x]);
    int r = x;
    while(r != p[r])
        r = p[r];
    int i = x, j;
    while(i != r) {
        j = p[i];
        p[i] = r;
        i = j;
    }
    return r;
}
double x[maxn], y[maxn];
int main() {
    int m;
    int T;
    while(cin >> T) {
        while(T--) {
            int n;
            cin >> n;
            for(int i = 1; i <= n; i++)
                p[i] = i;
            for(int i = 1; i <= n; i++) {
                cin >> x[i] >> y[i];
            }
            int flag = 0;
            int cnt = 0;
            for(int i = 1; i <= n; i++)
                for(int j = i + 1; j <= n; j++) {
                    ++cnt;
                    e[cnt].u = i;
                    e[cnt].v = j;
                    e[cnt].w =  sqrt(1.0 * (x[i] - x[j]) * (x[i] - x[j]) + 1.0 * (y[i] - y[j]) * (y[i] - y[j]));
                }
            double cost = 0.0;
            sort(e + 1, e + cnt + 1, cmp);
            for(int i = 1; i <= cnt; i++) {
                int x = find(e[i].u);
                int y = find(e[i].v);
                if(x == y)
                    continue;
                if(e[i].w<10||e[i].w>1000)
                    continue;
                cost += e[i].w * 100;
                p[x] = y;
                flag++;
            }
            if(flag == n-1)
                printf("%.1lf\n", cost);
            else cout << "oh!" << endl;
        }
    }
    return 0;
}