Codeforces 598E：Chocolate Bar

E. Chocolate Bar

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have a rectangular chocolate bar consisting of n × m single squares. You want to eat exactly k squares,
so you may need to break the chocolate bar.

In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length.

For example, if you have a chocolate bar consisting of 2 × 3 unit squares then you can break it horizontally and get two 1 × 3 pieces
(the cost of such breaking is 32 = 9),
or you can break it vertically in two ways and get two pieces: 2 × 1 and 2 × 2 (the
cost of such breaking is 22 = 4).

For several given values n, m and k find
the minimum total cost of breaking. You can eat exactly k squares of chocolate if after all operations of breaking there is a set
of rectangular pieces of chocolate with the total size equal to k squares. The remaining n·m - ksquares
are not necessarily form a single rectangular piece.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 40910) —
the number of values n, m and k to
process.

Each of the next t lines contains three integers n, m and k (1 ≤ n, m ≤ 30, 1 ≤ k ≤ min(n·m, 50)) —
the dimensions of the chocolate bar and the number of squares you want to eat respectively.

Output

For each n, m and k print
the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly ksquares.

Sample test(s)

input

4
2 2 1
2 2 3
2 2 2
2 2 4

output

5
5
4
0

Note

In the first query of the sample one needs to perform two breaks:

• to split 2 × 2 bar into two pieces of 2 × 1 (cost
  is 22 = 4),
• to split the resulting 2 × 1 into two 1 × 1 pieces
  (cost is 12 = 1).

In the second query of the sample one wants to eat 3 unit squares. One can use exactly the same strategy as in the first query of the sample.

一个n*m大的巧克力，你要吃k个单元的巧克力。每次切分都会有切的那条边长度平方的代价。问最小代价。

自己DP题目做得太少。

从最简单状况考虑，一块巧克力被切分，就分成了两块。这样从小到大，考虑各种情况(切的哪条边、两块中每一块吃多少个巧克力)递推。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

#define INF 1e9+7

int dp[32][32][52];

int main()
{
    //freopen("i.txt","r",stdin);
    //freopen("o.txt","w",stdout);
    int i, j, k, h, m;
    for (i = 0; i <= 30; i++)
    {
        for (j = 0; j <= 30; j++)
        {
            for (k = 0; k <= 50; k++)
            {
                if (k == i*j || k == 0)
                {
                    dp[i][j][k] = 0;
                }
                else
                {
                    dp[i][j][k] = INF;
                }
                for (h = 0; h <= k; h++)
                {
                    for (m = 1; m < j; m++)
                        dp[i][j][k] = min(dp[i][j][k], dp[i][m][h] + dp[i][j - m][k - h] + i*i);
                    for (m = 1; m < i; m++)
                        dp[i][j][k] = min(dp[i][j][k], dp[m][j][h] + dp[i - m][j][k - h] + j*j);
                }
            }
        }
    }

    scanf("%d", &k);
    while (k--)
    {
        scanf("%d%d%d", &i, &j, &h);
        printf("%d\n", dp[i][j][h]);
    }
    //system("pause");
    return 0;
}

看到好多时间特别短的是深搜，然后把当前值记录下来，不仅仅是这道题，上一道深搜的题目也是，有的时候为了减少时间，不一定全部预处理，用到哪个再求哪个，然后把求到的记录下来，为了下次询问使用。