学步园

An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b. 
Write a program that: finds the minimal number of elements in a set containing at least two different integers from each interval.

The first line of the input contains the number of intervals n, 1 <= n <= 10000. Each of the following n lines contains two integers a, b separated by a single space, 0 <= a < b <= 10000. They are the beginning and the end of an interval.

Output the minimal number of elements in a set containing at least two different integers from each interval.

4
3 6
2 4
0 2
4 7

4

CEOI 1997

0 1 2 3 4 5 6 7
-----           (0 to 2)
    -----       (2 to 4) 
      -------   (3 to 6)
        ------- (4 to 7)

数据如上，要找的就是一个集合，集合与数据的每个集合的交集至少有2个元素，求集合最少的数的个数。

贪心：先按照结尾数排序，第一个集合找末尾2个数，第二个集合，现根据前面判断已经有几个数在交集中了，如果够2个就跳到第三个集合，如果不够就从末尾找，往前补够2个，第三及以后的类推。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
    int a,b;
}p[10005];
int n,ans;
bool cmp(node x,node y)
{
    return x.b<y.b;
}
bool v[10005];
int main()
{
    while(scanf("%d",&n)!=EOF){
    for(int i=0;i<n;i++)scanf("%d%d",&p[i].a,&p[i].b);
    sort(p,p+n,cmp);
    memset(v,0,sizeof(v));
    v[p[0].b-1]=1;
    v[p[0].b]=1;
    ans=2;
    for(int i=1;i<n;i++)
    {
        int num=2;
        for(int j=p[i].a;j<=p[i].b;j++)
        {
            if(v[j]==1)num--;
        }
        if(num>0)
        while(num)
        {
            v[p[i].b-num+1]=1;
            ans++;
            num--;
        }
    }
    cout<<ans<<endl;}
    return 0;
}