Restaurant--贪心

一个关于老板想赚更多钱的问题。
贪心算法。
Restaurant
Time Limit:4000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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 ri each (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

运用到贪心算法，首先将结束时间排序，前一个的结束时间<后一个的开始时间。
代码如下：

#include <cstdio>#include <algorithm>using namespace std;struct n{    int str,end;}d[500000+11];bool cmp(n a,n b){    return a.end < b.end;}int main(){    int t;    while(~scanf("%d",&t))    {        for (int i=0;i<t;i++)            scanf ("%d %d",&d[i].str,&d[i].end);        sort(d,d+t,cmp);        int pos=0;        int ans=1;        for (int i=1;i<t;i++)        {            if (d[i].str>d[pos].end)            {                pos=i;                ans++;            }        }        printf("%d\n",ans);    }    return 0;}