Removing Columns

最新推荐文章于 2023-08-05 01:15:07 发布

原创最新推荐文章于 2023-08-05 01:15:07 发布 · 604 阅读

1 ·

CC 4.0 BY-SA版权

水题同时被 2 个专栏收录

21 篇文章

订阅专栏

uva

1 篇文章

订阅专栏

本文介绍了一种算法，该算法旨在通过移除表格中尽可能少的列来确保每一行按字母顺序排列，最终实现整个表格按行进行字典序排序。文章详细解释了输入输出格式，并提供了一个示例代码实现。

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 test(s)

Input

1 10
codeforces

Output

Input

4 4
case
care
test
code

Output

Input

5 4
code
forc
esco
defo
rces

Output

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <cmath>
#define LL long long
#define INF 0x3f3f3f3f

using namespace std;

char s[110][110];
bool vis[110];

int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        memset(vis,false,sizeof(vis));
        int cnt = 0;
        for(int i=0;i<n;i++)
            cin>>s[i];
        bool flag;
        for(int j=0;j<m;j++)
        {
            flag = false;
            for(int i=1;i<n;i++)
            {
                if(!vis[i] && s[i][j] < s[i-1][j])
                {
                    flag = true;
                    cnt++;
                    break;
                }
            }
            if(!flag)
            {
                for(int i=1;i<n;i++)
                {
                    if(s[i][j] > s[i-1][j])
                    {
                        vis[i] = true;
                    }
                }
            }
        }
        cout<<cnt<<endl;
    }
    return 0;
}