51nod 1133 不重叠的线段 (贪心，序列上的区间问题)

题意：

最多能选几条不重叠的线段

思路：

按R从小到大排序，维护一个最大的右端点

右端点最小的那个线段是必选的，可以贪心地证明

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
    
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;

int n;
struct node{
    int x, y;
}a[maxn];
bool cmp(node a, node b){
    //if(a.y==b.y)return a.x<b.x;
    return a.y<b.y;
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%d%d", &a[i].x,&a[i].y);
    }
    sort(a+1,a+1+n,cmp);
    int ans = 1;
    int r = a[1].y;
    for(int i = 1; i <= n; i++){
        if(a[i].x>=r){
            ans++;
            r = a[i].y;
        }
    }
    printf("%d",ans);
    return 0;
}