Selfish Grazing（貪心）

題目鏈接

題意：給你n個線段讓你求最多有多少個不相交的線段。

思路：把它看成會場安排問題

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=1e4+7;
const int N=2e5+10;
const int inf=0x7f7f7f7f;


ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

ll lcm(ll a,ll b)
{
    return a*(b/gcd(a,b));
}

template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-')
            op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op)
        x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0)
        x = -x, putchar('-');
    if(x >= 10)
        write(x / 10);
    putchar('0' + x % 10);
}
PII a[N];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&a[i].second,&a[i].first);
       // a[i].second+=a[i].first-1;
    }
    sort(a,a+n);
    int ans=0,low=a[0].first;
    for(int i=1;i<n;i++)
    {

        if(low<=a[i].second)
        {
            ans++; low=a[i].first;
        }
    }
    cout<<ans+1<<endl;
    return 0;
}