扩散

题目描述

思路

二分的是结果，使用并查集保存各个点之间是否连通
a为开始的点，b, c为目标点， 在经过了某个时刻时，a直接到不了b，a直接能到c，光通过这两个条件不能确定a间接到不了b， a可能通过c再到达b
并查集维护的就是目标点之间的连通性，不光要验证a能不能到达b，还要验证目标点内部能不能互相到达。

代码

#include <cstdio>
#include <cmath>

int n, inf = 0x7f7f7f7f;
struct Node {
	int a, b;
} m[55];
int fa[55], l, r, mid, ans, cnt, len;
int rec[55][55];
int dis(Node a, Node b) {
	return std::abs(a.a - b.a) + std::abs(a.b - b.b);
}

int get(int x) {
	if (x == fa[x]) return x;
	return get(fa[x]);
}
inline int read() {
	int s = 0, f = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
	return s * f;
}
int main() {
	n = read();
	for (int i = 1; i <= n; ++i) m[i].a = read(), m[i].b = read();
	for (int i = 1; i <= n; ++i) {
		for (int j = i + 1; j <= n; ++j) {
			rec[i][j] = dis(m[i], m[j]);
		}
	}
	l = 0, r = inf;
	while (l <= r) {
		mid = l + r >> 1, len = mid << 1;
		for (int i = 1; i <= n; ++i) fa[i] = i;
		for (int i = 1; i <= n; ++i) {
			for (int j = i + 1; j <= n; ++j) {
				int fax = get(i), fay = get(j);
				if (fax != fay) {
					if(rec[i][j] <= len) fa[fax] = fay;
				} 
			}
		}
		bool flag = true;
		for (int i = 2, j = get(1); i <= n && flag; ++i) {
			if (get(i) != j) flag = false;
		}
		if (flag) ans = mid, r = mid - 1;
		else l = mid + 1;
	}
	printf("%d", ans);
	return 0;
}