Minimal coverage(贪心区间覆盖)-白红宇

Minimal coverage(贪心区间覆盖)

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发布时间：2019-05-25

本文共 1756 字，大约阅读时间需要 5 分钟。

Given several segments of line (int the X axis) with coordinates [Li

, Ri

]. You are to choose the minimal

amount of them, such they would completely cover the segment [0, M].

Input

The first line is the number of test cases, followed by a blank line.

Each test case in the input should contains an integer M (1 ≤ M ≤ 5000), followed by pairs “Li Ri”

(|Li

|, |Ri

| ≤ 50000, i ≤ 100000), each on a separate line. Each test case of input is terminated by pair

‘0 0’.

Each test case will be separated by a single line.

Output

For each test case, in the first line of output your programm should print the minimal number of line

segments which can cover segment [0, M]. In the following lines, the coordinates of segments, sorted

by their left end (Li), should be printed in the same format as in the input. Pair ‘0 0’ should not be

printed. If [0, M] can not be covered by given line segments, your programm should print ‘0’ (without

quotes).

Print a blank line between the outputs for two consecutive test cases.

Sample Input

-1 0

-5 -3

2 5

0 0

-1 0

0 1

0 0

Sample Output

0 1

给你一个数m,然后给你几个区间,让你求这几个区间中能够覆盖[0,m]的最小个数.

明显的贪心.我们按着结尾的大小,由大到小去将区间排序…然后一个一个的去查询,并且随时更新区间.

代码如下:

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        using namespace std;struct node{	int start;	int end;}p[100005],q[100010];int cmp(const node &a,const node &b){	return a.end>b.end;}int s=0;int e;int main(){	int t;	scanf("%d",&t);	while(t--)	{		s=0;		scanf("%d",&e);		int cnt=0;		while(scanf("%d%d",&p[cnt].start,&p[cnt].end),p[cnt].end+p[cnt].start)		{			cnt++;		}		sort(p,p+cnt,cmp);		int ant=0;		int i;		while(s
        
         s) { s=p[i].end; q[ant++]=p[i]; break; } } if(i==cnt) break; } if(s