【23.58%】【code forces 321E】Ciel and Gondolas

time limit per test4 seconds

memory limit per test512 megabytes

inputstandard input

outputstandard output

Fox Ciel is in the Amusement Park. And now she is in a queue in front of the Ferris wheel. There are n people (or foxes more precisely) in the queue: we use first people to refer one at the head of the queue, and n-th people to refer the last one in the queue.

There will be k gondolas, and the way we allocate gondolas looks like this:

When the first gondolas come, the q1 people in head of the queue go into the gondolas.

Then when the second gondolas come, the q2 people in head of the remain queue go into the gondolas.

…

The remain qk people go into the last (k-th) gondolas.

Note that q1, q2, …, qk must be positive. You can get from the statement that and qi > 0.

You know, people don’t want to stay with strangers in the gondolas, so your task is to find an optimal allocation way (that is find an optimal sequence q) to make people happy. For every pair of people i and j, there exists a value uij denotes a level of unfamiliar. You can assume uij = uji for all i, j (1 ≤ i, j ≤ n) and uii = 0 for all i (1 ≤ i ≤ n). Then an unfamiliar value of a gondolas is the sum of the levels of unfamiliar between any pair of people that is into the gondolas.

A total unfamiliar value is the sum of unfamiliar values for all gondolas. Help Fox Ciel to find the minimal possible total unfamiliar value for some optimal allocation.

Input

The first line contains two integers n and k (1 ≤ n ≤ 4000 and 1 ≤ k ≤ min(n, 800)) — the number of people in the queue and the number of gondolas. Each of the following n lines contains n integers — matrix u, (0 ≤ uij ≤ 9, uij = uji and uii = 0).

Please, use fast input methods (for example, please use BufferedReader instead of Scanner for Java).

Output

Print an integer — the minimal possible total unfamiliar value.

Examples

input

5 2

0 0 1 1 1

0 0 1 1 1

1 1 0 0 0

1 1 0 0 0

1 1 0 0 0

output

0

input

8 3

0 1 1 1 1 1 1 1

1 0 1 1 1 1 1 1

1 1 0 1 1 1 1 1

1 1 1 0 1 1 1 1

1 1 1 1 0 1 1 1

1 1 1 1 1 0 1 1

1 1 1 1 1 1 0 1

1 1 1 1 1 1 1 0

output

7

input

3 2

0 2 0

2 0 3

0 3 0

output

2

Note

In the first example, we can allocate people like this: {1, 2} goes into a gondolas, {3, 4, 5} goes into another gondolas.

In the second example, an optimal solution is : {1, 2, 3} | {4, 5, 6} | {7, 8}.

题解

DP；

设f[i][j]表示前i艘船。装下j个人的最小不友好值。

f[i][j] = min(f[i-1][k]+w[k+1][j]);

其中w[i][j]表示从把第i个人到第j个人放在同一艘船上增加的不友好值。

这个不友好值可以用一个类似”矩阵前缀和”的东西弄出来;具体的看代码；

这里主要是f[i][j]这个转移的优化方法;

用到了四边形不等式;

想看证明的话转到这个地址:

http://www.cnblogs.com/vongang/archive/2013/01/21/2869315.html

大概就是说设s[i][j]为f[i][j]这个状态转移所需要的决策量。

如果满足BALABALBA就有s[i-1][j] < s[i][j] < s[i][j+1];

用这个就能把n^3的复杂度降低到n^2;

如果不用getchar输入会T

代码

#include <cstdio>
#include <cctype>
#include <cstring>

const int MAXN = 4010;
const int MAXM = 810;

int n, m,a[MAXN][MAXN],b[MAXN][MAXN],w[MAXN][MAXN];
int f[MAXM][MAXN];
int s[MAXM][MAXN];

void input(int &num)
{
    num = 0;
    char c;
    do
    {
        c = getchar();
    } while (!isdigit(c));
    while (isdigit(c))
    {
        num = num * 10 + c - '0';
        c = getchar();
    }
}

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    input(n); input(m);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            input(a[i][j]);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)//矩阵前缀和
            a[i][j] = a[i][j] + a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];
    for (int i = 1; i <= n; i++)
        for (int j = i; j <= n; j++)//后面加上重复减去的部分。除2就是花费
            w[i][j] = (a[j][j] - a[i - 1][j] - a[j][i - 1] + a[i - 1][i - 1]) / 2;
    memset(f, 127 / 3, sizeof(f));
    for (int i = 1; i <= n; i++)
        f[1][i] = w[1][i], s[1][n] = 0;
    for (int i = 2; i <= m; i++)
    {
        s[i][n + 1] = n;
        for (int j = n; j >= i; j--)
        {
            for (int k = s[i - 1][j]; k <= s[i][j + 1]; k++)
                if (f[i][j] > f[i - 1][k] + w[k + 1][j])
                    f[i][j] = f[i - 1][k] + w[k + 1][j], s[i][j] = k;//更改决策量。
        }
    }
    printf("%d\n", f[m][n]);
    return 0;
}