Codeforces 496C Removing Columns

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

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.

Rrond 283 Div1 的A题水题。

只要暴力判断每一列都符合升序，在进行标记处理那些严格升序的行。

最后被hack 了 ， 错在边判断边标记 ， 一旦中途遇到不符合升序的列，前面的标记没有消除到。

简直悲惨ya~

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
string s[N];
bool tag , vis[N] ;
int ans , n  , m ;

bool check( int j ){
    for( int i = 1 ; i < n ; ++i ){
        if( s[i][j] < s[i-1][j] && !vis[i] )
            return false;
    }
    return true;
}

void solve( int j ){
    for( int i = 1 ; i < n ; ++i )
        if( s[i][j] > s[i-1][j] ) vis[i] = true ;
}

int main(){
    cin >> n >> m ;
    for( int i = 0 ; i < n ; ++i )
        cin >> s[i];
    ans = 0 ; bool tag = 0 ;
    for( int i = 0 ; i < m ; ++i ) {
        tag = check(i);
        if( tag ) solve(i);
        else ans++;
    }
    cout << ans << endl ;
}