bzoj 1604: [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居

Description

了解奶牛们的人都知道，奶牛喜欢成群结队．观察约翰的N(1≤N≤100000)只奶牛，你会发现她们已经结成了几个“群”．每只奶牛在吃草的时候有一个独一无二的位置坐标Xi，Yi(l≤Xi，Yi≤[1．.10^9]；Xi，Yi∈整数．当满足下列两个条件之一，两只奶牛i和j是属于同一个群的：

1．两只奶牛的曼哈顿距离不超过C(1≤C≤10^9)，即lXi - xil+IYi - Yil≤C.

2．两只奶牛有共同的邻居．即，存在一只奶牛k，使i与k，j与k均同属一个群．

给出奶牛们的位置，请计算草原上有多少个牛群，以及最大的牛群里有多少奶牛

Input

第1行输入N和C，之后N行每行输入一只奶牛的坐标．

Output

仅一行，先输出牛群数，再输出最大牛群里的牛数，用空格隔开．

Sample Input

4 2

1 1

3 3

2 2

10 10

Line 1: A single line with a two space-separated integers: the

number of cow neighborhoods and the size of the largest cow

neighborhood.

Sample Output

2 3

OUTPUT DETAILS:

There are 2 neighborhoods, one formed by the first three cows and

the other being the last cow. The largest neighborhood therefore

has size 3.

Solution

每个点的坐标改为x′=x+y,y′=x−y，这样条件就变成了max(|x′1−x′2|,|y′1−y′2|)<=c。

于是将新坐标按x排序，从左往右扫，维护一个队列，其中的元素x坐标之差不超过c，然后每次新加一个点时考虑其y坐标在队列中的点的前驱与后继，如果y坐标之差不超过c则用并查集合并。

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 100005;

int n, c;

struct Point{
int x, y, id;
Point() {}
Point(int a, int b, int c) : x(a), y(b), id(c) {}
bool operator < (const Point &p) const {
return x < p.x;
}
};

int ans;
Point p[MAXN];
int fa[MAXN], cnt[MAXN];
multiset<pair<int, int> > s;

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

void un(int a, int b) {
int x = find(a), y = find(b);
if (x != y) ans--;
fa[x] = y;
}

int main() {
scanf("%d %d", &n, &c);
int a, b;
for (int i = 1; i <= n; i++) {
scanf("%d %d", &a, &b);
p[i] = Point(a + b, a - b, i);
}
sort(p + 1, p + n + 1);
for (int i = 1; i <= n; i++)
fa[i] = i;
ans = n;
int l = 1;
s.insert(make_pair(p[1].y, p[1].id));
for (int i = 2; i <= n; i++) {
// cout << p[i].x << ' ' << p[i].y << endl;
while (l < i && p[i].x - p[l].x > c) {
s.erase(s.find(make_pair(p[l].y, p[l].id)));
l++;
}
set<pair<int, int> >::iterator it = s.lower_bound(make_pair(p[i].y, p[i].id));
if (it != s.end() && it -> first - p[i].y <= c) un(p[i].id, it -> second);
if (it != s.begin() && p[i].y - (--it) -> first <= c) un(p[i].id, it -> second);
s.insert(make_pair(p[i].y, p[i].id));
}
int ma = 0;
for (int i = 1; i <= n; i++) {
cnt[find(i)]++;
ma = max(ma, cnt[find(i)]);
}
printf("%d %d\n", ans, ma);
return 0;
}