Problem:
Score : 800 points

Problem Statement

Mr. Takahashi has in his room an art object with H rows and W columns, made up of H \times W blocks.
Each block has a color represented by a lowercase English letter (a-z).
The color of the block at the i-th row and j-th column is c_{i,j}.

Mr. Takahashi would like to dismantle the object, finding it a bit kitschy for his tastes.
The dismantling is processed by repeating the following operation:

Choose one of the W columns and push down that column one row.
The block at the bottom of that column disappears.

Each time the operation is performed, a cost is incurred.
Since blocks of the same color have a property to draw each other together, the cost of the operation is the number of the pairs of blocks (p, q) such that:

The block p is in the selected column.

The two blocks p and q are horizontally adjacent (before pushing down the column).

The two blocks p and q have the same color.

Mr. Takahashi dismantles the object by repeating the operation H \times W times to get rid of all the blocks.
Compute the minimum total cost to dismantle the object.

Constraints

1 ≤ H ≤ 300

2 ≤ W ≤ 300

All characters c_{i,j} are lowercase English letters (a-z).

Partial Score

In test cases worth 300 points, W = 3.

Input

The input is given from Standard Input in the following format:

H W
c_{1,1}c_{1,2}..c_{1,W}
c_{2,1}c_{2,2}..c_{2,W}
:
c_{H,1}c_{H,2}..c_{H,W}

Output

Print the minimum total cost to dismantle the object.

Sample Input 1

2 3
rrr
brg

Sample Output 1

2

For example, the total cost of 2 can be achieved by performing the operation as follows and this is the minimum value.

Sample Input 2

6 3
xya
xya
ayz
ayz
xaz
xaz

Sample Output 2

0

The total cost of 0 can be achieved by first pushing down all blocks of the middle column, then all of the left column, and all of the right column.

Sample Input 3

4 2
ay
xa
xy
ay

Sample Output 3

0

The total cost of 0 can be achieved by the following operations:

pushing down the right column one row;

pushing down the left column one row;

pushing down all of the right column;

and pushing down all of the left column.

Sample Input 4

5 5
aaaaa
abbba
ababa
abbba
aaaaa

Sample Output 4

24

Sample Input 5

7 10
xxxxxxxxxx
ccccxxffff
cxxcxxfxxx
cxxxxxffff
cxxcxxfxxx
ccccxxfxxx
xxxxxxxxxx

Sample Output 5

130

Solution in Rust (1.15.1) (judge verdict: Accepted)

// https://atcoder.jp/contests/code-festival-2016-qualc/tasks/codefestival_2016_qualC_d
//
#![allow(unused_imports)]
use std::io::*;
use std::io::Write;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

// Input macros.
// Original by tanakh: https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };

    ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, [ next / $t:tt ]) => {
        {
            let len = read_value!($iter, usize);
            (0..len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
        }
    };

    ($iter:expr, switch) => {
        {
            let ty = read_value!($iter, i32);
            if ty == 1 {
                vec![ty, read_value!($iter, i32), read_value!($iter, i32)]
            } else if ty == 2 {
                vec![ty, read_value!($iter, i32)]
            } else {
                vec![ty, read_value!($iter, i32)]
            }
        }
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_macros)]
macro_rules! read_line {
    ($t:tt) => {
        {
            let mut s = String::new();
            std::io::stdin().read_line(&mut s).unwrap();
            s.trim_right().parse::<$t>().unwrap()
        }
    }
}

#[allow(unused_macros)]
macro_rules! dvec {
    ($t:expr ; $len:expr) => {
        vec![$t; $len]
    };

    ($t:expr ; $len:expr, $($rest:expr),*) => {
        vec![dvec!($t; $($rest),*); $len]
    };
}

#[allow(unused_macros)]
macro_rules! ifv {
    ($t:expr, $a:expr, $b: expr) => {
        if $t { $a } else { $b }
    }
}

#[allow(unused_macros)]
macro_rules! fill {
    ($t:expr, $v:expr) => {
        for i in 0..$t.len() {
            $t[i] = $v;
        }
    };
}

#[allow(unused_macros)]
macro_rules! join {
    ($t:expr, $glue:expr) => {
        $t.into_iter().map(|w| w.to_string()).collect::<Vec<_>>().join($glue)
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        println!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

// ===

const INF: i64 = 1e12 as i64;

fn solve(a: &Vec<char>, b: &Vec<char>) -> i64 {
    let n = a.len();
    let mut cost = dvec!(0; n+1, n+1);
    let mut dp = dvec!(INF; n+1, n+1);
    dp[n][n] = 0;

    for i in 1..n+1 {
        for j in 1..n+1 {
            cost[i][j] = cost[i-1][j-1] + ifv!(a[i-1] == b[j-1], 1, 0);
        }
    }

    for i in (0..n+1).rev() {
        for j in (0..n+1).rev() {
            if i >= 1 {
                dp[i-1][j] = min(dp[i-1][j], dp[i][j] + cost[i][j]);
            }
            if j >= 1 {
                dp[i][j-1] = min(dp[i][j-1], dp[i][j] + cost[i][j]);
            }
        }
    }
    dp[0][0]
}

fn main() {
    input! {
        h: usize, w: usize,
        table: [chars; h]
    };

    let mut data = vec![vec![' '; h]; w];
    for i in 0..w {
        for j in 0..h {
            data[i][j] = table[j][i];
        }
    }
    let mut cost = 0;
    for i in 0..w-1 {
        cost += solve(&data[i], &data[i+1]);
    }
    println!("{}", cost);
}