题内话

同 P1696，又是一道简单的暴力水题。

第一层循环枚举被解雇的奶牛的编号，第二层循环枚举除了被解雇的奶牛，其他奶牛的工作时间，第三重循环从从工作时间的左端点枚举到右端点，打个标记，再把所有打上了标记的时间点的个数都统计一下，答案就是统计出的结果里面的最大值。

详见代码：

#include<iostream>
using namespace std;

int n, ans = -1e7;
struct node{
	int x, y;
}a[105];
bool vis[10005];

int main(){
    cin >> n;
    for (int i = 1; i <= n; i++){
    	cin >> a[i].x >> a[i].y;
	}
	for (int i = 1; i <= n; i++){
		int cnt = 0;
		for (int j = 0; j <= 1000; j++) vis[j] = 0;
		for (int j = 1; j <= n; j++){
			if (j == i) continue;
			for (int k = a[j].x; k < a[j].y; k++){
				vis[k] = 1;
			}
		}
		for (int j = 0; j <= 1000; j++) if (vis[j]) cnt++;
		ans = max(ans, cnt);
	}
	cout << ans;
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int n, a[101], b[101], t[1001], s[1001];
int main(){
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> n;
    int mi=1001,mx=0;
    memset(t, 0, sizeof(t));
    for (int i = 1; i <= n; i++){
        cin >> a[i] >> b[i];a[i]++;
        mi = min(mi, a[i]);
        mx = max(mx, b[i]);
        for (int j = a[i]; j <= b[i]; j++)t[j]++;
    }
    int S = 0;for (int i = 1; i <= 1000; i++)S += (t[i] > 0);
    s[0] = 0;
    for (int i = 1; i <= 1000; i++)s[i] = s[i - 1] + (t[i]==1);
    int ans=1001;
    for (int i = 1; i <= n; i++)ans = min(ans, s[b[i]] - s[a[i]-1]);
    cout << S - ans << endl;
    return 0;
}