POJ1716 Integer Intervals

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13984		Accepted: 5943

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

Source

CEOI 1997

题目和POJ1201相同。

由于这次区间内的数的个数固定为2，所以可以使用贪心解法。将区间按右端点从小到大排序，每次能不加点就不加，必须加的话就加在区间最右面，并累计答案数。

 /**/
 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int mxn=;
 struct area{
     int l,r;
 }a[mxn];
 int n;
 int ans,f,s;
 int cmp(area a,area b){
     if(a.r!=b.r)return a.r<b.r;
     return a.l<b.l;
 }
 int main(){
     scanf("%d",&n);
     int i,j;
     for(i=;i<=n;i++)scanf("%d%d",&a[i].l,&a[i].r);
     sort(a+,a+n+,cmp);
     ans=;f=a[].r;s=a[].r-;
     for(i=;i<=n;i++){
         if(f<a[i].l){
             ans+=;
             f=a[i].r;
             s=a[i].r-;
             continue;
         }
         if(s<a[i].l){
             ans++;
             if(f>a[i].r-)    s=a[i].r-;
             else s=f;
             f=a[i].r;
             continue;
         }
         if(f>a[i].r)f=a[i].r;
     }
     printf("%d\n",ans);
     return ;
 }