Ada, Bertrand and Charles often argue over which TV shows to watch, and to avoid some of their fights they have finally decided to buy a video tape recorder. This fabulous, new device can record kk different TV shows simultaneously, and whenever a show recorded in one the machine's kk slots ends, the machine is immediately ready to record another show in the same slot.
The three friends wonder how many TV shows they can record during one day. They provide you with the TV guide for today's shows, and tell you the number of shows the machine can record simultaneously. How many shows can they record, using their recording machine? Count only shows that are recorded in their entirety.
Input Format
The first line of input contains two integers nn, kk (1 le k < n le 100 000)(1≤k<n≤100000). Then follow nn lines, each containing two integers x_i, y_ixi,yi, meaning that show ii starts at time x_ixi and finishes by time y_iyi. This means that two shows iiand jj, where y_i = x_jyi=xj, can be recorded, without conflict, in the same recording slot. You may assume that 0 le x_i < y_i le 1 000 000 0000≤xi<yi≤1000000000.
Output Format
The output should contain exactly one line with a single integer: the maximum number of full shows from the TV guide that can be recorded with the tape recorder.
样例输入1
3 1 1 2 2 3 2 3
样例输出1
2
样例输入2
4 1 1 3 4 6 7 8 2 5
样例输出2
3
样例输入3
5 2 1 4 5 9 2 7 3 8 6 10
样例输出3
3
题目来源
Nordic Collegiate Programming Contest 2015
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 #include <string> 6 #include <string> 7 #include <map> 8 #include <cmath> 9 #include <set> 10 #include <algorithm> 11 using namespace std; 12 const int N=1e5+9; 13 struct Node 14 { 15 int s,e; 16 }node[N]; 17 bool cmp(Node a,Node b) 18 { 19 return a.e<b.e;//按终止时间从小到大 20 } 21 int n,k; 22 multiset<int>se;//可以存储等值元素 23 /* 24 1 8 25 2 9 26 5 10 27 7 10 28 _____ 29 就会有10 10 (k==2) 30 */ 31 multiset<int>::iterator it;//注意：：写在前面 32 int main() 33 { 34 scanf("%d%d",&n,&k); 35 for(int i=0;i<n;i++ ) scanf("%d%d",&node[i].s,&node[i].e); 36 sort(node,node+n,cmp); 37 se.clear(); 38 for(int i=0;i<k;i++) se.insert(0); 39 int ans=0; 40 for(int i=0;i<n;i++) 41 { 42 it=se.upper_bound(node[i].s); 43 if(it==se.begin()) continue; 44 it--;//第一个>node[i].s的数的前面的数一定是小于node[i].sb_type 45 // 并且一定是和node[i].s的差值最小的（贪心），把更早结束的留给开始时间比较小的 46 se.erase(it);//播放完了，就要更新掉/ 47 se.insert(node[i].e); 48 ans++; 49 } 50 printf("%d ",ans); 51 return 0; 52 }