| 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.
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
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。
1 /**/ 2 #include<iostream> 3 #include<cstdio> 4 #include<cmath> 5 #include<cstring> 6 #include<algorithm> 7 using namespace std; 8 const int mxn=50000; 9 struct area{ 10 int l,r; 11 }a[mxn]; 12 int n; 13 int ans,f,s; 14 int cmp(area a,area b){ 15 if(a.r!=b.r)return a.r<b.r; 16 return a.l<b.l; 17 } 18 int main(){ 19 scanf("%d",&n); 20 int i,j; 21 for(i=1;i<=n;i++)scanf("%d%d",&a[i].l,&a[i].r); 22 sort(a+1,a+n+1,cmp); 23 ans=2;f=a[1].r;s=a[1].r-1; 24 for(i=2;i<=n;i++){ 25 if(f<a[i].l){ 26 ans+=2; 27 f=a[i].r; 28 s=a[i].r-1; 29 continue; 30 } 31 if(s<a[i].l){ 32 ans++; 33 if(f>a[i].r-1) s=a[i].r-1; 34 else s=f; 35 f=a[i].r; 36 continue; 37 } 38 if(f>a[i].r)f=a[i].r; 39 } 40 printf("%d ",ans); 41 return 0; 42 }