Removing Columns - CodeForces 496 C 水题

C. 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

abcdedfghijk

 

we obtain the table:

acdefghjk

 

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 10codeforces

output

0

input

4 4casecaretestcode

output

2

input

5 4codeforcescodeforces

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.

题意：问至少删除几列，使得n个字符串按照字典序排列。

思路：从左到右扫，每次记录当前这个字符串的前i个字符是否和上一个相同，如果相同的话，那么就后面跟的必须字典序，否则的话就可以随意了。

AC代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>using namespace std;typedef long long ll;char s[1010][1010];int dp[2][1010];int main(){    int T,t,n,m,i,j,k,ans=0,a,b,c;    bool flag;    scanf("%d%d",&n,&m);    for(i=1;i<=n;i++)       scanf("%s",s[i]+1);    for(i=2;i<=n;i++)       dp[0][i]=1;    a=0,b=1;    for(j=1;j<=m;j++)    {        i=1;        flag=true;        while(i<=n)        {            dp[b][i]=0;            for(i++;i<=n && dp[a][i]==1;i++)            {                if(s[i][j]==s[i-1][j])                  dp[b][i]=1;                else if(s[i][j]>s[i-1][j])                  dp[b][i]=0;                else                {                    flag=false;                    break;                }            }            if(!flag)              break;        }        if(flag)          swap(a,b);        else          ans++;    }    printf("%d\n",ans);}