BNU Concentric Rings

There are several different concentric rings on the ground. Some of them may overlap. In Figure 1, there are 3 rings: blue, green and red. The red one is just above the green one. The problem is to remove minimum number of rings so that no two of the remaining overlap.

Figure 1Input The input consists of multiple test cases. Each test case starts with a positive integer N (<=10000) which represents the number of rings. The next N lines each line contains two positive integers which represents the inner radius and outer radius respectively.
Output For each test case, output the minimum number of rings to remove.
Sample Input: 3
1 2
3 6
4 5
Sample Output: 1

Source
CYJJ's Funny Contest #1, Killing in Seconds

這題和杭電上有個看電視的那個一樣，簡單貪心題，這個題首先輸入有n個環，接下來每行都是環的內半徑和外半徑，然後求滿足不能有重疊所需要刪除幾個環。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

struct Node
{
    int inner,outer;
}a[10010];

bool cmp(Node a, Node b)
{
    return a.outer < b.outer;
}

int main()
{
    int n,i,j,sum;
    while(scanf("%d",&n)!=EOF)
    {
        sum = 0;
        for(i = 0; i < n; i++)
        {
            scanf("%d%d",&a[i].inner,&a[i].outer);
        }
        sort(a,a+n,cmp);
        j = 0;
        for(i = 1; i < n; i++)
        {
            if(a[i].inner < a[j].outer)
            {
                sum++;
                a[i].inner = a[i].outer = 0;
            }
            else
            {
                j = i;
            }
        }
        printf("%d\n",sum);
    }

    return 0;
}