P1522 [USACO2.4] 牛的旅行 Cow Tours

思路

对于每个连通块求出任意两点距离的最大值 \(max[u]\)，然后再 \(O(n^2)\) 枚举任意两个连通块连接起来，答案就是两个连通块的最大值 \(max\) 加上中间连接的边的长度。

代码

#include <bits/stdc++.h>

using namespace std;
using PDD = pair<double, double>;

const int N = 160;

int fa[N];

int find(int x) {
	if (x == fa[x]) return x;
	return fa[x] = find(fa[x]);
}

void merge(int x, int y) {
	int fx = find(x), fy = find(y);
	if (fx == fy) return;
	fa[fx] = fy;
}

int n;
PDD a[N];
double dis[N][N], maxd[N];
double zjd[N];

double gdis(PDD a, PDD b) {
	return sqrt(((a.first - b.first) * (a.first - b.first)) + ((a.second - b.second) * (a.second - b.second)));
}

int main() {	
	cin >> n;
	for (int i = 1; i <= n; i++) fa[i] = i;
	for (int i = 1; i <= n; i++) cin >> a[i].first >> a[i].second;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			char c;
			cin >> c;
			if (c == '1') {
				dis[i][j] = gdis(a[i], a[j]);
				merge(i, j);
			}
			else dis[i][j] = 1e18;
		}
	}
	
	for (int k = 1; k <= n; k++) {
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				if (dis[i][j] > dis[i][k] + dis[k][j]) {
					dis[i][j] = dis[i][k] + dis[k][j];
				}
			}
		}
	}
	
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			int fx = find(i), fy = find(j);
			if (fx == fy && i != j) {
				zjd[fx] = max(dis[i][j], zjd[fx]);
			}
		}
	}
	
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (i != j && find(i) == find(j) && dis[i][j] < 1e17) maxd[i] = max(maxd[i], dis[i][j]);
		}
	}
	
	double ans = 1e18;
	
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (find(i) == find(j)) continue;
			ans = min(ans, max(max(zjd[find(i)], zjd[find(j)]), maxd[i] + maxd[j] + gdis(a[i], a[j])));
		}
	}
	printf("%.6lf", ans);
	return 0;
}