Restaurant

Restaurant
Time Limit:4000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
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Description

A restaurant received n orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the i-th order is characterized by two time values — the start time li and the finish time ri (li ≤ ri).

Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept?

No two accepted orders can intersect, i.e. they can't share even a moment of time. If one order ends in the moment other starts, they can't be accepted both.

Input

The first line contains integer number n (1 ≤ n ≤ 5·105) — number of orders. The following n lines contain integer values li and rieach (1 ≤ li ≤ ri ≤ 109).

Output

Print the maximal number of orders that can be accepted.

Sample Input

Input
2
7 11
4 7
Output
1
Input
5
1 2
2 3
3 4
4 5
5 6
Output
3
Input
6
4 8
1 5
4 7
2 5
1 3
6 8
Output
2
/*
题意:给你餐厅n组客人就餐时间(起始时间,终止时间)如果时间有重叠的话,那么两组客人不能同时用餐,问餐厅最多接纳多少组客人

初步思路:按照终止时间排序,然后贪心

*/
#include <bits/stdc++.h>
using namespace std;
struct node{
    int Frist,End;
    bool operator <(const node &other) const{
        return End<other.End;
    }
}fr[500005];
int n;
int res=0;
int last=0;
void init(){
    res=0;
    last=0;
}
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF){
        init();
        for(int i=0;i<n;i++){
            scanf("%d%d",&fr[i].Frist,&fr[i].End);
        }
        sort(fr,fr+n);
        for(int i=0;i<n;i++){
            if(fr[i].Frist>last){
                res++;
                last=fr[i].End;
            }
        }
        printf("%d\n",res);
    }
    return 0;
}

 

posted @ 2017-03-30 17:41  勿忘初心0924  阅读(341)  评论(0编辑  收藏  举报