acwing-387. 北极网络

考点

  • 最小生成树

题解

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#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e2 + 50, inf = 0x3f3f3f3f;
int n, s;
int d[maxn], a[maxn][maxn];
bool v[maxn];

struct node {
int x, y;
} p[maxn];

int f(int i, int j) {
int x1 = p[i].x, x2 = p[j].x, y1 = p[i].y, y2 = p[j].y;
return (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
}

void init() {
cin >> s >> n;
for (int x, y, i = 1; i <= n; ++i) {
cin >> x >> y;
p[i].x = x, p[i].y = y;
}
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
if (i == j)
a[i][j] = 0x3f3f3f3f;
else
a[i][j] = a[i][j] = f(i, j);
}

void prim() {
memset(v, 0, sizeof(v)), memset(d, 0x3f, sizeof(d));
d[1] = 0;
for (int k = 1; k < n; ++k) {
int x = 0;
for (int i = 1; i <= n; ++i)
if (!v[i] && (!x || d[x] > d[i])) x = i;
v[x] = 1;
for (int y = 1; y <= n; ++y)
if (!v[y] && d[y] > a[x][y]) {
d[y] = a[x][y];
}
}
sort(d + 1, d + 1 + n);
cout << fixed << setprecision(2) << sqrt(d[n - s + 1]) << endl;
}

int main() {
int t;
cin >> t;
while (t--) init(), prim();
return 0;
}

思路

最小生成树的裸题。

题目要让前哨站连通,先用prim求出最小生成树,再对最小生成树排序;

卫星能免s - 1条边,那么答案就是sqrt(d[n - s + 1])