poj 1089 贪心之区间覆盖问题

最新推荐文章于 2025-06-25 12:40:00 发布

柠檬与茉莉

最新推荐文章于 2025-06-25 12:40:00 发布

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分类专栏：贪心文章标签： acm 区间覆盖

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博客详细解析了POJ 1089题目，通过贪心策略解决区间覆盖问题。内容包括题目的输入输出格式、样例输入输出及对应的解题代码，适合ACM竞赛爱好者和算法学习者参考。

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There is given the series of n closed intervals [ai; bi], where i=1,2,...,n. The sum of those intervals may be represented as a sum of closed pairwise non−intersecting intervals. The task is to find such representation with the minimal number of intervals. The intervals of this representation should be written in the output file in acceding order. We say that the intervals [a; b] and [c; d] are in ascending order if, and only if a <= b < c <= d.
Task
Write a program which:
reads from the std input the description of the series of intervals,
computes pairwise non−intersecting intervals satisfying the conditions given above,
writes the computed intervals in ascending order into std output

Input

In the first line of input there is one integer n, 3 <= n <= 50000. This is the number of intervals. In the (i+1)−st line, 1 <= i <= n, there is a description of the interval [ai; bi] in the form of two integers ai and bi separated by a single space, which are respectively the beginning and the end of the interval,1 <= ai <= bi <= 1000000.

Output

The output should contain descriptions of all computed pairwise non−intersecting intervals. In each line should be written a description of one interval. It should be composed of two integers, separated by a single space, the beginning and the end of the interval respectively. The intervals should be written into the output in ascending order.

Sample Input

Sample Output

1 4
5 10

思路：用最少的线段来覆盖所给的区间，属于贪心问题；贪心策略：将所有区间按左端点值从小到大排序，然后遍历选择。因为要用最少的线段，所以就必须选择最大的长度，我们是希望区间重叠的个数越多越好~这样就可以让线段长度最大，而数量达到最小，对吧！

附上代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <algorithm>
using namespace std;
struct node
{
  int x,y;
}a[50100];
int cmp(node a,node b)
{
 if(a.x!=b.x)
    return a.x<b.x;
 return a.y<b.y;
}
int s[50100],e[50100];
int main()
{
 int n;
 while(cin>>n)
 {
   int cnt=0;
   for(int i=0;i<n;i++)
     cin>>a[i].x>>a[i].y;
   sort(a,a+n,cmp);
   s[cnt]=a[0].x;
   e[cnt]=a[0].y;
   for(int i=1;i<n;i++)
   {
     if(a[i].x>e[cnt])
     {
       cnt++;
       s[cnt]=a[i].x;
       e[cnt]=a[i].y;
     }
     else 
     {
       if(a[i].y>e[cnt])
       e[cnt]=a[i].y;
     }
   }
  for(int i=0;i<=cnt;i++)
  {
    cout<<s[i]<<" "<<e[i]<<endl;
  }
 }
 return 0;
}