Minimal coverage(贪心区间覆盖)

最新推荐文章于 2022-06-20 14:50:34 发布

starlet_kiss

最新推荐文章于 2022-06-20 14:50:34 发布

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文章标签：贪心区间更新思维

本文链接：https://blog.youkuaiyun.com/starlet_kiss/article/details/84591730

本文介绍了一种基于贪心策略的区间覆盖算法，旨在从给定的多个线段中选择最少数量的线段，以完全覆盖指定的目标区间[0,M]。通过详细解析算法流程和步骤，包括输入输出格式、样例输入输出，以及具体实现代码，帮助读者深入理解算法原理及其应用。

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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
2
1
-1 0
-5 -3
2 5
0 0
1
-1 0
0 1
0 0
Sample Output
0
1
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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
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<e)
		{
			for(i=0;i<cnt;i++)
			{
				if(p[i].start<=s&&p[i].end>s)
				{
					s=p[i].end;
					q[ant++]=p[i];
					break;
				}
			}
			if(i==cnt) break;
		}
		if(s<e) printf("0\n");
		else 
		{
			printf("%d\n",ant);
			for(i=0;i<ant;i++)
			{
				printf("%d %d\n",q[i].start,q[i].end);
			}
		}
		if(t) cout<<endl; 
	}
}

努力加油a啊,(^o)/~