Integer Intervals（贪心）

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12123		Accepted: 5129

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：在数轴上给n个区间，区间上的点均是整数，再在数轴上取x个点构成的点集V使得V与上述每个区间的交集至少包含两个交点，输出满足题意的x的最小值。
思路：可以贪心，先按每个区间的有端点升序排序，
初始化：计数器 sum = 2,集合V的前两个元素selem,telem为第一个区间的最后两个整数；（selem和telem在整个循环中作为集合V的最后两个元素，所以必要时更新其值）
for：
　　若第i+1个区间包含selem和telem，不需要做任何改变；
　　若第i+1个区间只包含telem,更新selem和telem,sum加1；
　　若第i+1个区间不包含selem和telem,更新selem和telem,sum加2；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

struct node
{
    int s,t;
}L[10010];
int cmp(const struct node &a, const struct node &b)
{
    return a.t < b.t;
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 0; i < n; i++)
        scanf("%d %d",&L[i].s,&L[i].t);
    sort(L,L+n,cmp);
    int sum,selem,telem;
    sum = 2;
    selem = L[0].t-1;
    telem = L[0].t;
    for(int i = 1; i < n;)
    {
        if(selem >= L[i].s && selem <= L[i].t && telem >= L[i].s && telem <= L[i].t)
            i++;
        else if(selem < L[i].s && telem >= L[i].s && telem <= L[i].t)
        {
            selem = telem;
            telem = L[i].t;
            sum++;
            i++;
        }
        else if(selem < L[i].s && telem < L[i].s)
        {
            selem = L[i].t-1;
            telem = L[i].t;
            sum += 2;
            i++;
        }
    }
    printf("%d
",sum);
    return 0;
}