Towers CodeForces - 229D

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 hi. 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 hi (1 ≤ hi ≤ 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.

Example

5
8 2 7 3 1      

3      

3
5 2 1      

2      

題意： 給出n個正整數，進行若幹個操作，使得序列非減，求最少的操作次數；

            操作：

                    每次可以選擇兩個相鄰的數合并為一個；

解法：

         (1) dp[i][j]表示 前i個整數合并成非減序列的最小代價，且最後一段區間為j->i

         (2) 枚舉最後一段合并的區間；

dp(i)表示使得前i個塔美麗的最小操作次數，sum(i)表示前i座塔的字首和，last(i)表示使得前i個塔美麗操作次數最小的情況下，最右側一座塔最小的塔高。

那麼就有狀态轉移方程:dp(i)=min{dp(j)+i-j+1},sum(i)-sum(j)>=last(j).

#include <cstdio>
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;
                if(last[i] > sum[i]-sum[j]) last[i] = sum[i]-sum[j];
            }
        }
    }
    printf("%d\n",dp[n]);
    return 0;
}      

原文位址http://blog.sina.com.cn/s/blog_140e100580102wkl5.html