CodeForces-1132C Painting the Fence

题目链接

https://vjudge.net/problem/CodeForces-1132C

题面

Description

You have a long fence which consists of \(n\) sections. Unfortunately, it is not painted, so you decided to hire \(q\) painters to paint it. \(i\)-th painter will paint all sections \(x\) such that \(l_i \le x \le r_i\).

Unfortunately, you are on a tight budget, so you may hire only \(q - 2\) painters. Obviously, only painters you hire will do their work.

You want to maximize the number of painted sections if you choose \(q - 2\) painters optimally. A section is considered painted if at least one painter paints it.

Input

The first line contains two integers \(n\) and \(q\) (\(3 \le n, q \le 5000\)) — the number of sections and the number of painters availible for hire, respectively.

Then \(q\) lines follow, each describing one of the painters: \(i\)-th line contains two integers \(l_i\) and \(r_i\) (\(1 \le l_i \le r_i \le n\)).

Output

Print one integer — maximum number of painted sections if you hire \(q - 2\) painters.

Examples

Input

7 5
1 4
4 5
5 6
6 7
3 5

Output

7

Input

4 3
1 1
2 2
3 4

Output

2

Input

4 4
1 1
2 2
2 3
3 4

Output

3

题意

墙长为n，q个工人，每个工人固定粉刷一个区间，区间可能重叠，现在要去掉两个工人，求剩下q-2个工人最多粉刷多少墙

题解

我们可以\(n^2\)枚举去掉的两个工人，然后O(1)查询剩下了多少墙

O(1)查询的方法是：先预处理出每一段墙有多少工人刷，然后处理出每一个工人刷的只有他自己刷的墙的长度，然后预处理出只有两个工人刷的墙的区间的前缀和，然后对于枚举的两个工人，刷过的墙的总数首先要减去两个工人刷的墙的部分中只有一个人刷的部分，因为去掉这个工人这个墙就没人刷了，如果两个工人的覆盖区间有重叠，假设重叠区间\([L,R]\)，要减去的区间就是\([L,R]\)中只有两个工人刷的部分，因为去掉这两个工人这段墙就没人刷了

结果取最大值即可

AC代码

#include <bits/stdc++.h>
#define N 5050
using namespace std;
struct seg {
	int l, r;
} a[N];
int pre[N];
int pre1[N];
int pre2[N];
int max(int a, int b) {
	return a > b ? a : b;
}
int min(int a, int b) {
	return a < b ? a : b;
}
int main() {
	int n, q;
	scanf("%d%d", &n, &q);
	int sum = 0;
	for (int i = 1; i <= q; i++) {
		scanf("%d%d", &a[i].l, &a[i].r);
		pre[a[i].l]++; pre[a[i].r + 1]--;
	}
	for (int i = 1; i <= n; i++) {
		pre[i] += pre[i - 1];
	}
	for (int i = 1; i <= n; i++) {
		if (pre[i]) sum++;
	}
	for (int i = 1; i <= n; i++) {
		if (pre[i] == 2) pre1[i]++;
	}
	for (int i = 1; i <= n; i++) {
		pre1[i] += pre1[i - 1];
	}
	for (int i = 1; i <= q; i++) {
		for (int j = a[i].l; j <= a[i].r; j++) {
			if (pre[j] == 1) pre2[i]++;
		}
	}
	int L, R;
	int ans = 0;
	for (int i = 1; i <= q; i++) {
		for (int j = i + 1; j <= q; j++) {
			L = max(a[i].l, a[j].l);
			R = min(a[i].r, a[j].r);
			int tmp = sum;
			if (R >= L) {
				tmp = tmp - (pre1[R] - pre1[L - 1]);
			}
			tmp -= pre2[i] + pre2[j];
			ans = max(tmp, ans);
		}
	}
	printf("%d\n", ans);
	return 0;
}