CF-45D. Event Dates（贪心）

最新推荐文章于 2021-02-21 20:30:25 发布

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本文介绍了一种解决历史事件日期选择问题的算法。该算法帮助确定一系列著名历史事件的确切发生日期，在已知每个事件可能发生的时间段的情况下，确保任意两天内不会发生超过一个事件。通过选择每个时间段的最早日期，并调整后续时间段的开始日期来避免冲突。

On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [l_i, r_i] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers l_i and r_ieach (1 ≤ l_i ≤ r_i ≤ 10⁷) — the earliest acceptable date and the latest acceptable date of the i-th event.

Output

Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.

Sample test(s)

input
3
1 2
2 3
3 4

output
1 2 3 

input
2
1 3
1 3

output
1 2 

思路：每次选择左边界最小，次之右边界最小，选择此区间的左边界值l作为此区间代表,并删去本区间，然后提升剩余区间的左边界使之大于那个区间的代表l，重复此步骤即可

#include<iostream>
#include<cstring>
using namespace std;
const int mm=110;
class node
{
  public:int l,r;
}f[mm];
bool vis[mm];
int ans[mm];
bool ok(int a,int b)
{
  if(f[a].l<f[b].l)return 1;
  else if(f[a].l==f[b].l&&f[a].r<f[b].r)return 1;
  return 0;
}
int main()
{
  int n;
  while(cin>>n)
  {
    for(int i=0;i<n;i++)
    {
      cin>>f[i].l>>f[i].r;vis[i]=0;
    }
    int k;
    for(int i=0;i<n;i++)
    { k=-1;
      for(int j=0;j<n;j++)
        if(!vis[j]&&(k==-1||ok(j,k)))///选择区间左界最小
          k=j;
        int w=ans[k]=f[k].l;
        ++w;vis[k]=1;
      for(int j=0;j<n;j++)///提升剩余区间的左界
        if(w>f[j].l)
        f[j].l=w;
    }
    for(int i=0;i<n;i++)
    cout<<ans[i]<<" ";cout<<"\n";
  }
}