POJ - 1716 Integer Intervals

Integer Intervals

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 14913		Accepted: 6310

Description

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.

Input

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

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

Sample Input

4
3 6
2 4
0 2
4 7

Sample Output

4

题意：

        输入n个区间，让求一个集合，每个区间至少有两个整数包含在这个集合内，求集合最少包含几个元素。

题解：

        贪心。先将集合按照右边界从小到大排序。取两个变量first，second（first < second）分别为当前区间取的两个值。优先右取，这样可以保证first和second可以包含在尽量多的区间内。不断维护first和second的值。如果当前区间的左边界大于second，就需要在本区间内重新取两个值，sum+2；如果左边界小于等于second且大于first，则将second赋给first，second取右边界（保证first < second），sum+1；如果左边界小于first，则不需要改变first和second的值。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int left,right;
}p[10005];
bool cmp(node a,node b)
{
    return a.right<b.right;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
            scanf("%d%d",&p[i].left,&p[i].right);
        sort(p,p+n,cmp);
        int first=p[0].right-1,second=p[0].right,sum=2;
        for(int i=1;i<n;i++)
        {
            if(p[i].left>second)
            {
                sum+=2;
                second=p[i].right;
                first=second-1;
            }
            else if(p[i].left<=second && p[i].left>first)
            {
                sum++;
                first=second;
                second=p[i].right;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}