答案显然具有单调性, 二分答案, 判断时可以用并查集维护, 注意两个点可以连通的条件 ((|xi-xj|+|yi-yj|+1)/2 <= mid)。
时间复杂度(O(nlogn*log_2T)).
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define FOR(a, b, c) for(int (a) = b; (a) <= (c); (a)++)
const int mx = 55;
struct node{
int x, y;
}a[mx];
int n;
int fat[mx];
inline void reset() { FOR(i, 1, n) fat[i] = i; }
inline int find(int x) { return x == fat[x] ? x : fat[x] = find(fat[x]); }
int dist(int i, int j) {
return (abs(a[i].x - a[j].x) + abs(a[i].y - a[j].y) + 1)/2;
}
inline bool check(int x) {
reset();
FOR(i, 1, n-1)
FOR(j, i+1, n) {
if(dist(i, j) <= x) {
fat[find(i)] = find(j);
}
}
int cnt = 0;
FOR(i, 1, n) if(fat[i] == i) cnt++;
return cnt == 1;
}
int main() {
cin >> n;
FOR(i, 1, n) scanf("%d %d", &a[i].x, &a[i].y);
int l = 0, r = 1e9;
FOR(i, 1, 100) {
int mid = l + (r-l)/2;
if(check(mid)) r = mid;
else l = mid;
}
//mid~r为可行的, 所以答案为r, l错误
cout << r;
return 0;
}