CodeForces 496C 字符串阵处理

Description

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 andm (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 Input

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

Hint

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 islexicographically larger than t if they are not equal and the character following the largest common prefix ofs and t (the prefix may be empty) ins is alphabetically larger than the corresponding character oft.

题意:把一行与下一行进行比较,此行字典序必须小于等于下一行,    否则就要删除不满足条件的那一列的所以字母   ,使剩下的字符阵满足上面条件.

---

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<algorithm>

using namespace std;

int n,m,x;
char s[1000][1000];
int vlei[10000];
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int num=0;
        for(int i=0; i<n; i++)
        {
            scanf("%s",s[i]);
        }

        for(int j=0; j<m; j++)
            for(int i=1; i<n; i++)
            {
                if(s[i-1][j]>s[i][j]&&strcmp(s[i-1],s[i])>0)  
                {
                    for(int k=0; k<n; k++)
                    {
                        s[k][j]='0';
                    }
                    num++;
                    break;
                }
            }

//        for(int i=0; i<n; i++)
//        {
//            printf("%s\n",s[i]);
//        }
        printf("%d\n",num);
    }
}