2021“MINIEYE杯”中国大学生算法设计超级联赛（8）- 1004（Ink on paper）

补题：​​https://acm.hdu.edu.cn/showproblem.php?pid=7058​​

每两个点之间连边，构建一个完全图，权值为两点之间的距离，跑一边prim算法，找到最小生成树边中最大的边输出即可!!!

​​code​​

#include <iostream>
#include <cstring>
using namespace std;

typedef long long ll;

const int N = 5005;

int n, x[N], y[N];
bool st[N];
ll dis[N], g[N][N];

void prim(){

    ll res = 0;
    dis[0] = 0;

    for (int i = 0; i < n; ++i) {

        int t = -1;
        for (int j = 0; j < n; ++j) {
            if(!st[j] && (t == -1 || dis[j] < dis[t])){
                t = j;
            }
        }

        st[t] = true;
        if(i) res = max(res, dis[t]);

        for (int j = 0; j < n; j++) {

            dis[j] = min(dis[j], g[t][j]);
        }
    }

    printf("%lld\n", res);
}

int main(){

    int t;
    scanf("%d", &t);

    while (t--){

        memset(dis, 0x3f, sizeof(dis));
        memset(st, 0, sizeof(st));

        scanf("%d", &n);

        for (int i = 0; i < n; ++i) {

            scanf("%d%d", &x[i], &y[i]);
        }

        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {

                ll d1 = (x[i] - x[j]);
                ll d2 = (y[i] - y[j]);

                ll c = d1*d1 + d2*d2;

                g[i][j] = c, g[j][i] = c;
            }
        }

        prim();
    }

    return 0;
}