CodeForces 496C Removing Columns

C. Removing Columns

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

abcdedfghijk

 

we obtain the table:

acdefghjk

 

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 10codeforces

output

0

input

4 4casecaretestcode

output

2

input

5 4codeforcescodeforces

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.

题目大意：给定一个n乘m的字符矩阵，删除一些列后的按行字典序排列。求最少删除多少列。

解题思路：模拟排除，从左往右扫描每一列，直到找到符合行字典序的列作为首列，然后从上往下扫描，字母不同后面的序列不考虑，字母相同要判断后面的各列是否满足行字典序。

解题过程：模拟容易超时，程序做了多处优化才AC，但效率不佳。wa点：注意考虑某列被排除后其后的各列要重新判断是否满足条件，细节注意n和m易混淆。

代码如下：

#include <cstdio>#include <cstring>char s[102];int a[102][102]; int w[102];   //标记该列是否被排除int n,m,ans;bool p(int x){    for(int i=1;i<n;i++)        if(a[x][i-1]>a[x][i])            return false;    return true;}void pp(int x,int u,int v){if(a[x][u]<a[x][v]&&w[x]==0)//满足字典序且字母不相等，不用进入后面的判断，直接返回。return;    else if(a[x][u]>a[x][v]&&w[x]==0)//不满足字典序，ans增加，该列标记，不进入下一次判断    {        ans++;        w[x]=1;      }x++;if(x<m)pp(x,u,v);return ; }int main(){    freopen ("in.txt","r",stdin);    int i,j;    ans=0;memset(w,0,sizeof(w));    scanf("%d%d",&n,&m);    for(i=0;i<n;i++)    {        scanf("%s",s);        for(j=0;j<m;j++)            a[j][i]=s[j]-'0';    }    for(i=0;i<m;i++)    {        if(!p(i))         {            ans++;            continue;        }        else if(i<m-1) // 满足条件，作为首列{int t=m-i-1;while(t--){for(j=1;j<n;j++)if(a[i][j-1]==a[i][j]) //字母相同，判断之后各列是否满足条件pp(i+1,j-1,j);}}break;    }    printf("%d\n", ans);    return 0;}