Ink on paper HDU - 7058

题意：

给出n个墨水的初始位置，每秒向外扩展0.5cm，显示一个圆圈，问所有墨水连接起来需要多长时间

题解：

很明显，在完全图中找最小生成树，并记录最小生成树中最长的边
数据N<=5000,因为是完全图，边很多，所以跑prim肯定没错

代码：

#include <bits/stdc++.h>
using namespace std;
int read() {int tot = 0, fh = 1;char c = getchar();while ((c < '0') || (c > '9')) {if (c == '-')fh = -1;c = getchar();}while ((c >= '0') && (c <= '9')) {tot = tot * 10 + c - '0';c = getchar();}return tot * fh;
}
const int maxn = 5010;
const long long inf = 9 * 1e18;
struct node {long long x, y;
} a[maxn];
long long mp[maxn][maxn], f[maxn];
long long ans;
int i, j, k;
int n, mi, t;
long long len(int x, int y) {return (a[x].x - a[y].x) * (a[x].x - a[y].x) + (a[x].y - a[y].y) * (a[x].y - a[y].y);
}
int opt, T;
int main() {//  freopen("data.in","r",stdin);//  freopen("data.out","w",stdout);T = read();for (opt = 1; opt <= T; opt++) {n = read();ans = 0;for (i = 1; i <= n; i++) {a[i].x = read();a[i].y = read();}for (i = 1; i <= n; i++) {mp[i][i] = 0;for (j = i + 1; j <= n; j++) {mp[i][j] = len(i, j);mp[j][i] = mp[i][j];}}for (i = 1; i <= n; i++)f[i] = mp[1][i];f[1] = inf;for (k = 1; k < n; k++) {mi = 1;for (i = 2; i <= n; i++)if (f[mi] > f[i])mi = i;t = 1;for (i = 1; i <= n; i++)if ((f[i] == inf) && (mp[t][mi] > mp[i][mi]))t = i;ans = max(ans, mp[t][mi]);f[mi] = inf;for (i = 1; i <= n; i++)if (f[i] != inf)f[i] = min(f[i], mp[mi][i]);}printf("%lld\n", ans);}return 0;
}