POJ

Integer Intervals

Time Limit: 1000MS Memory Limit: 10000KTotal 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

43 62 40 24 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;}