CF 496C(Removing Columns-贪心取字典序)

最小化删除列数以形成字典序递增表格

最新推荐文章于 2019-11-05 09:32:16 发布

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最新推荐文章于 2019-11-05 09:32:16 发布

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分类专栏：字典序贪心

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解决了一个关于优化表格列删除数量的问题，目标是通过最少的列删除操作使得表格中行的字典序形成递增顺序。通过贪心策略逐列判断并决定是否保留，最终计算所需的最小删除次数。

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
0

input
4 4
case
care
test
code

output
2

input
5 4
code
forc
esco
defo
rces

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：有若干长度相等的字符串依次排列。问至少删除若干列，才能使字符串从上到下字典序单调不降。

显然，第一行如果不满足字典序一定要删。

如果满足字典序（如果前缀不同可分开考虑），那么取这一段一定不劣

（因为对于每个前缀一致的一段，

eg1：

AAABxxxx。。

AAACxxxx。。

可以看到由于前缀不同，后面x可乱取。

eg2:

AAABxxxx。。

前缀一样，对后面不影响还是要保证字典序。

∴贪心，从左到右每列能留就留。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
char a[MAXN][MAXN];
int pre[MAXN];
int main()
{
//	freopen("Columns.in","r",stdin);
//	freopen(".out","w",stdout);
	int n,m;
	cin>>n>>m;
	For(i,n) scanf("%s",a[i]+1),pre[i]=1;
	int ans=0;
	For(j,m) 
	{
		bool flag=0;
		Fork(i,2,n) 
		{
			if (i==pre[i]) continue;
			if (pre[i]<i)
			{
				if (a[i][j]<a[i-1][j]) {flag=1;break;}
			}	
			
		}
		ans+=flag;
		
		if (flag==0)
		{
			Fork(i,2,n)
			{
				if (i==pre[i]) continue;
				if (pre[i]<i&&a[i][j]!=a[i-1][j]) pre[i]=i;
				else pre[i]=pre[i-1];
			}
		} 
		
	} 
	cout<<ans<<endl;
	
	return 0;
}