CF-45D. Event Dates(贪心)

D. Event Dates
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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 [li, ri] (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 li and rieach (1 ≤ li ≤ ri ≤ 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";
  }
}




posted @ 2013-03-05 13:52  剑不飞  阅读(277)  评论(0编辑  收藏  举报