数轴上有n个闭区间[ai,bi]。取尽量少的点,使得每个区间内都至少有一个点(不同区间内含的点可以是同一个)。
例题如下:
N one dimensional kingdoms are represented as intervals of the form [ai , bi] on the real line.
A kingdom of the form [L, R] can be destroyed completely by placing a bomb at a point x on the real line if L
≤ x ≤ R.
Your task is to determine minimum number of bombs required to destroy all the one dimensional kingdoms.
Input
First line of the input contains T denoting number of test cases.
For each test case, first line contains N denoting the number of one dimensional kingdoms.
For each next N lines, each line contains two space separated integers ai and bi.
Output
For each test case , output an integer denoting the minimum number of bombs required.
Constraints
Subtask 1 (20 points) : 1 ≤ T ≤ 100 , 1 ≤ N ≤ 100 , 0 ≤ ai ≤ bi ≤500
Subtask 2 (30 points) : 1 ≤ T ≤ 100 , 1 ≤ N ≤ 1000 , 0 ≤ ai ≤ bi ≤ 1000
Subtask 3 (50 points) : 1 ≤ T ≤ 20 , 1 ≤ N ≤ 105, 0 ≤ ai ≤ bi ≤ 2000
Example
Input:
1
3
1 3
2 5
6 9
Output:
2
附上链接:http://www.codechef.com/JAN15/problems/ONEKING
退役之后无聊,就刷了刷CC,发现自己真是生锈了。这题我一开始看,这很简单啊,直接按照左端点来排序,然后进行选择,但是,WA了一上午,最后别人告诉我这是典型的区间取点问题,搜了一下,发现,这贪心就是必须要按照右端点进行排序,才有用。这样就一下AC了。真是年轻。
附上代码:
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn=1e5+10;
struct node
{
int l,r;
} A[maxn];
bool cmp(node k1,node k2)
{
if(k1.r==k2.r)
{
return k1.l<k2.l;
}
else
{
return k1.r<k2.r;
}
}
int main()
{
int t,n,cnt,end1;
scanf("%d",&t);
while(t--)
{
cnt = 0;
end1 = -1;
scanf("%d",&n);
for(int i=0; i<n; i++)
scanf("%d%d",&A[i].l,&A[i].r);
sort(A,A+n,cmp);
for(int i=0; i<n; i++)
{
if(end1 < A[i].l)
{
end1 = A[i].r;
cnt++;
}
}
printf("%d\n",cnt);
}
return 0;
}