51Nod 1133 （贪心）

题意：
X轴上有N条线段，每条线段有1个起点S和终点E。最多能够选出多少条互不重叠的线段。

思路：
根据区间右端点排序之后贪心

代码：

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <map>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <sstream>
#define pb push_back
#define X first
#define Y second
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pii pair<int,int>
#define qclear(a) while(!a.empty())a.pop();
#define lowbit(x) (x&-x)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define mst(a,b) memset(a,b,sizeof(a))
#define cout3(x,y,z) cout<<x<<" "<<y<<" "<<z<<endl
#define cout2(x,y) cout<<x<<" "<<y<<endl
#define cout1(x) cout<<x<<endl
#define IOS std::ios::sync_with_stdio(false)
#define SRAND srand((unsigned int)(time(0)))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
using namespace std;
const double PI=acos(-1.0);
const int INF=0x3f3f3f3f;
const ll mod=998244353;
const double eps=1e-5;
const int maxn=2005;
const int maxm=10005;
struct node{
    int l,r;
    bool sel;
    bool operator < (const node b)const{
        if(r==b.r)return l<b.l;
        return r<b.r;
    }
};
node a[10005];
int n;
void solve() {
    sd(n);
    for(int i=0;i<n;i++){
        sdd(a[i].l,a[i].r);
    }
    sort(a,a+n);
    int pre=a[0].l;
    a[0].sel=1;
    for(int i=1;i<n;i++){
        if(a[i].l<pre){
            a[i].sel=0;
        }else{
            pre=a[i].l;
            a[i].sel=1;
        }
    }
    pre=a[0].r;
    int ans=1;
    for(int i=1;i<n;i++){
        if(!a[i].sel)continue;
        if(a[i].l<pre)continue;
        pre=a[i].r;
        ans++;
    }
    printf("%d\n",ans);
    return ;
}
int main() {
#ifdef LOCAL
    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
#else
    //    freopen("","r",stdin);
    //    freopen("","w",stdout);
#endif
    solve();
    return 0;
}