Removing Columns - CodeForces 496 C 水题

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本文探讨如何通过最少次数删除列，使包含n个字符串的矩形表格按照字典序排列。通过逐列扫描比较，实现高效排序，提供算法实现代码。

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.

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);
}