Codeforces Round #283(Div.2) C. Removing Columns 폭력

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.
 
제목을 보니까 한참 봤는데...

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 100001

const int inf=0x7fffffff;   //    

int n,m,res;

char c[2000][2000];

string s[2000];

int main()

{

    cin >> n >> m;

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

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

            cin >> c[i][h];

    bool b=0;

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

    {

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

        {

            if( s[h-1]+c[h-1][i] > s[h]+c[h][i] )

            {

                b=1;

                break;

            }

        }

        if(!b)

        {

            for(int h=0;h<n;h++)

                s[h]+=c[h][i];

        }    b=0;

    }

    cout << m-s[0].length();

    return 0;

}