学步园

Mr. and Mrs. Smith are going to the seaside for their holiday. Before they start off, they need to choose a hotel. They got a list of hotels from the Internet, and want to choose some candidate hotels which are cheap and close to the seashore. A candidate hotel
M meets two requirements: 

1. Any hotel which is closer to the seashore than M will be more expensive than M. 
2. Any hotel which is cheaper than M will be farther away from the seashore than M.

There are several test cases. The first line of each test case is an integer N (1 <= N <= 10000), which is the number of hotels. Each of the following N lines describes a hotel, containing two integers D and C (1 <= D, C <= 10000). D means the distance from
the hotel to the seashore, and C means the cost of staying in the hotel. You can assume that there are no two hotels with the same D and C. A test case with N = 0 ends the input, and should not be processed.

For each test case, you should output one line containing an integer, which is the number of all the candidate hotels.

5
300 100
100 300
400 200
200 400
100 500
0

2

Beijing 2005

水题，刚刚看时头晕了，。。。。。。

把价钱与距离看作是点的x，y坐标，对点做横竖两个轴，左下方（包括轴上）没其他点，那么这个点就符合要求了！

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n;
struct node
{
    int m,l;
}p[10005];
bool cmp(node a,node b)
{
    if(a.m==b.m)return a.l<b.l;
    else return a.m<b.m;
}
int main()
{
    while(scanf("%d",&n)&&n)
    {
        for(int i=0;i<n;i++)scanf("%d%d",&p[i].l,&p[i].m);
        sort(p,p+n,cmp);
        int key=1000000,ans=0;
        for(int i=0;i<n;i++)
        {
            if(p[i].l<key)
            {
                ans++;
                key=p[i].l;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}