Removing Columns

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table

abcd

edfg

hijk

we obtain the table:

acd

efg

hjk

A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.

Input

The first line contains two integers — n and m (1 ≤ n, m ≤ 100).

Next n lines contain m small English letters each — the characters of the table.

Output

Print a single number — the minimum number of columns that you need to remove in order to make the table good.

Sample test(s)

Input

1 10

codeforces

Output

0

Input

4 4

case

care

test

code

Output

2

Input

5 4

code

forc

esco

defo

rces

Output

4

Note

In the first sample the table is already good.

In the second sample you may remove the first and third column.

In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).

Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.

判断每一列的字符是不是符合,如果符合则判断如果上一行的字符大余这一行,则标记一下,因为后面的无论是否符合,此时的字典序已将符合.

#include <map>

#include <set>

#include <list>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm>

using namespace std;

#define eps 1e-9

#define LL long long

#define PI acos(-1.0)

#define INF 0x3f3f3f3f

#define CRR fclose(stdin)

#define CWW fclose(stdout)

#define RR freopen("input.txt","r",stdin)

#pragma comment(linker,"/STACK:102400000")

#define WW freopen("output.txt","w",stdout)

const int Max =  160000;

char s[110][110];

bool vis[110][110];

int n,m;

bool Judge(int j)

{

    for(int i=1;i<n;i++)

    {

        if(!vis[i][i+1])

        {

            if(s[i][j]>s[i+1][j])

            {

                return true;

            }

        }

    }

    for(int i=1;i<n;i++)

    {

        if(!vis[i][i+1])

        {

            if(s[i][j]<s[i+1][j])

            {

                vis[i][i+1]=true;

            }

        }

    }

    return false;

}

int main()

{

    scanf("%d %d",&n,&m);

    for(int i=1;i<=n;i++)

    {

        scanf("%s",s[i]);

    }

    int ans=0;

    memset(vis,false,sizeof(vis));

    for(int i=0;i<m;i++)

    {

        if(Judge(i))

        {

            ans++;

        }

    }

    printf("%d\n",ans);

    return 0;

}

版权声明：本文为博主原创文章，未经博主允许不得转载。