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

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#codeforces

本文介绍了一种通过移除特定列使矩形表格按字典序排列的方法，并提供了一个示例性的C语言程序实现，该程序能有效地计算出使表格行按字母顺序排列所需的最小操作次数。

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

Input

4 4
case
care
test
code

Output

Input

5 4
code
forc
esco
defo
rces

Output

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.

这种题用bfs肯定要爆内存，所以没办法爆搜

仔细想想，如果遇到不合法的，一定要去掉，用一个数组标记第j列是否要去掉，然后比较字典序就行了

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

bool vis[110];
char str[110][110];

int main()
{
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
		for (int i = 0; i < n; ++i)
		{
			scanf("%s", str[i]);
		}
		int ans = 0;
		memset (vis, 0, sizeof(vis));
		while (1)
		{
			bool flag = false;
			for (int i = 1; i < n; ++i)
			{
				for (int j = 0; j < m; ++j)
				{
					if (vis[j] || str[i][j] == str[i - 1][j])
					{
						continue;
					}
					if (str[i][j] > str[i - 1][j])
					{
						break;
					}
					if (str[i][j] < str[i - 1][j])
					{
						flag = true;
						vis[j] = 1;
						ans++;
					}
				}
			}
			if (!flag)
			{
				break;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}