[Codeforces496C] Removing Columns

描述 Description

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.

输入格式 InputFormat

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.

输出格式 OutputFormat

Print a single number — the minimum number of columns that you need to remove in order to make the table good.

样例输入 SampleInput

4 4
case
care
test
code

样例输出 SampleOutput

2

Codeforces 496C

代码 Code

从左到右枚举每一列判断。

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn=105;
int i,j,t,n,m,l,r,k,z,y,x;
char a[maxn][maxn];
bool b[maxn];
int ans;
inline bool check(int s)
{
    int i,j,t;
    for (i=1;i<=n;i++) if (a[i][s]<a[i-1][s] && !b[i]) return false;
    return true;
}
int main()
{
    scanf("%d%d",&n,&m);
    for (i=1;i<=n;i++) scanf("%s",a[i]);
    memset(b,false,sizeof(b));
    for (ans=0,i=0;i<m;i++)
    {
        if (check(i))
        {
            for (j=1;j<=n;j++) if (a[j][i]>a[j-1][i]) b[j]=true;
        }
        else ans++;
    }
    printf("%d\n",ans);
    return 0;
}