AT_abc014_3 题解

思路

看到输入是区间，我们想到用差分数组来做

第l到r项都加x，c[l]加x，c[r]减x就可以了

每项为前面所有数的和

AC 代码

#include<bits/stdc++.h>
using namespace std;

#define ll long long

ll n,a,b,answer;

ll c[1000010];//差分数组

int main(){
    cin>>n;
    for(ll i = 1; i <= n; i++)
    {
        cin>>a>>b;
        c[a]++;
        c[b + 1]--;
    }
    answer = c[0];
    for(ll i = 1; i <= 1e6; i++)
    {
        c[i] += c[i - 1];
        answer = max(answer,c[i]);
    }
    cout<<answer<<endl;
    return 0;
}