寒假刷題３7: Codeforces Round #622 (Div. 2)C1. Skyscrapers (easy version)

題目鏈接:

Codeforces Round #622 (Div. 2)C1. Skyscrapers (easy version)

題目解析:

通過分析題目可以知道,最終情況一定是一個倒V字形

分別讓每一個點作爲最高點,記錄當前可建造的最高高度maxhigh,同時向左右兩邊遍歷,更新maxhigh並求出以此時的答案,最後所有答案取最大值即可

時間複雜度O(n^2)

AC代碼:

#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
#define ll long long

ll n,m[1010],a[1010];
int main()
{
   cin>>n;
   ll id,ans=0;
   for(int i=1;i<=n;i++) cin>>m[i];
   for(int i=1;i<=n;i++)
   {
       ll temp=m[i],maxx=m[i];
       //cout<<temp<<" ";
       for(int j=i-1;j>=1;j--)
       {
           maxx=min(maxx,m[j]);
           temp+=maxx;
       }
        //cout<<temp<<" ";
       maxx=m[i];
       for(int j=i+1;j<=n;j++)
       {
           maxx=min(maxx,m[j]);
           temp+=maxx;
       }
       //cout<<temp<<endl;
       if(temp>ans) ans=temp,id=i;

   }
   //cout<<ans<<endl;
   ll maxx=m[id];
   a[id]=m[id];
   for(int i=id-1;i>=1;i--)
   {
       maxx=min(maxx,m[i]);
       a[i]=maxx;
   }
   maxx=m[id];
   for(int i=id+1;i<=n;i++)
   {
       maxx=min(maxx,m[i]);
       a[i]=maxx;
   }
   for(int i=1;i<=n;i++)
    cout<<a[i]<<" ";
}