Ink on paper HDU - 7058

題意：

給出n個墨水的初始位置，每秒向外擴展0.5cm，顯示一個圓圈，問所有墨水連接起來需要多長時間

題解：

很明顯，在完全圖中找最小生成樹，并記錄最小生成樹中最長的邊
數據N<=5000,因為是完全圖，邊很多，所以跑prim肯定沒錯

代碼：

#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; }

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