CodeForces 229D Towers

最新推荐文章于 2020-09-10 14:27:48 发布

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最新推荐文章于 2020-09-10 14:27:48 发布

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本文探讨了如何通过最少的操作步骤使城市中按照特定顺序排列的塔楼达到非递减高度序列的问题。详细介绍了动态规划方法，通过计算每个塔楼的最低高度并确定合并操作的最优策略。

原题传送门：http://codeforces.com/problemset/problem/229/D

The city of D consists of n towers, built consecutively on a straight line. The height of the tower that goes i-th (from left to right) in the sequence equals h_i. The city mayor decided to rebuild the city to make it beautiful. In a beautiful city all towers are are arranged in non-descending order of their height from left to right.

The rebuilding consists of performing several (perhaps zero) operations. An operation constitutes using a crane to take any tower and put it altogether on the top of some other neighboring tower. In other words, we can take the tower that stands i-th and put it on the top of either the (i - 1)-th tower (if it exists), or the (i + 1)-th tower (of it exists). The height of the resulting tower equals the sum of heights of the two towers that were put together. After that the two towers can't be split by any means, but more similar operations can be performed on the resulting tower. Note that after each operation the total number of towers on the straight line decreases by 1.

Help the mayor determine the minimum number of operations required to make the city beautiful.

Input

The first line contains a single integer n (1 ≤ n ≤ 5000) — the number of towers in the city. The next line contains n space-separated integers: the i-th number h_i (1 ≤ h_i ≤ 10⁵) determines the height of the tower that is i-th (from left to right) in the initial tower sequence.

Output

Print a single integer — the minimum number of operations needed to make the city beautiful.

Sample test(s)

Input

5
8 2 7 3 1

Output

Input

3
5 2 1

Output

2

分析：

dp(i)表示对前i座塔进行操作后形成非递减序列所需要的最小操作步数

last(i)表示在dp(i)最小的前提下，第i座塔的最低高度。

易知，将[i,j]区间内的塔合并成一座需要j-i步

于是我们得到dp方程：dp(i)=max{dp(j)+i-j+1| j<i, h(j+1)+h(j+2)+...+h(i-1)+h(i) <= last(j)}

注意，我们在递推的过程中要同时记录、更新last(j)的值来确保得到最优解。

#include <cstdio>
using namespace std;
int dp[5010],sum[5010],last[5010];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++){
        int a;
        scanf("%d",&a);
         sum[i] = sum[i-1]+a;
         dp[i] = last[i] = 1<<30;
     }
     for(int i = 1;i <= n;i++){
         for(int j = 0;j < i;j++){
             if(sum[i]-sum[j] >= last[j] && dp[i] >= dp[j]+i-j-1){
                 dp[i] = dp[j]+i-j-1;
                 last[i] =sum[i]-sum[j]>last[i]?last[i]:sum[i]-sum[j];
             }
         }
     }
     printf("%d\n",dp[n]);
     return 0;
 }