Codeforces 496 C Removing Columns【思维】

最新推荐文章于 2018-07-10 10:46:43 发布

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本文介绍了一种算法问题，目标是最少的操作次数让矩阵的每一行按字典序排列。通过逐列检查并标记需要移除的列来解决该问题。

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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.

Examples

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.

题目大意：

给你n个字符串，每个字符串保证长度为m，每次操作可以去掉一列，问最少去掉多少次能够使得第i个字符串比第i-1个字符串字典序小（也可以相等）。

思路：

1、我们从第一列开始判断，一直判断到最后一列，首先，（假设当前判断到了第i列）如果当前列有第j+1个字符串比第j个字符串上的字母要小，那么这一列是一定要删除的，对应设定vis【i】，表示第i列需要删除，那么如果出现了这种情况，对应vis【i】=1；

2、那么如果这一列没有上述情况，那么对应这一列上边如果有：第j+1行上的这个字符比第j行上的这个字符大，那么第j行这个字符串一定满足了字典序要求，那么我们对应设定b【j】，表示第j行的字符串已经满足了字典序，之后的列的判断这个字符串可以不参与判断了，那么如果出现了这种情况，对应b【j】=1；

3、那么我们在判断第j列的时候，如果b【j】==1，那么这个字符串我们就不需要判断了。ans=vis【i】为1的个数。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
char a[150][150];
int vis[150];
int b[150];
int n,m;
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(vis,0,sizeof(vis));
        memset(b,0,sizeof(b));
        for(int i=0;i<n;i++)
        {
            scanf("%s",a[i]);
        }
        for(int i=0;i<m;i++)
        {
            int judge=0;
            for(int j=0;j<n-1;j++)
            {
                if(b[j]==1)continue;
                if(a[j+1][i]<a[j][i])judge=1;
            }
            if(judge==1)vis[i]=1;
            else
            {
                for(int j=0;j<n-1;j++)
                {
                    if(a[j+1][i]>a[j][i])b[j]=1;
                }
            }
        }
        int output=0;
        for(int i=0;i<m;i++)
        {
            if(vis[i]==1)output++;
        }
        printf("%d\n",output);
    }
}