C. Remembering Strings

Time Limit: 20 Sec Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/543/problem/C

Description

You have multiset of n strings of the same length, consisting of lowercase English letters. We will say that those strings are easy to remember if for each string there is some position i and some letter c of the English alphabet, such that this string is the only string in the multiset that has letter c in position i.

For example, a multiset of strings {"abc", "aba", "adc", "ada"} are not easy to remember. And multiset {"abc", "ada", "ssa"} is easy to remember because:

the first string is the only string that has character c in position 3;
the second string is the only string that has character d in position 2;
the third string is the only string that has character s in position 2.

You want to change your multiset a little so that it is easy to remember. For a_ij coins, you can change character in the j-th position of the i-th string into any other lowercase letter of the English alphabet. Find what is the minimum sum you should pay in order to make the multiset of strings easy to remember.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 20) — the number of strings in the multiset and the length of the strings respectively. Next n lines contain the strings of the multiset, consisting only of lowercase English letters, each string's length is m.

Next n lines contain m integers each, the i-th of them contains integers a_i1, a_i2, ..., a_im (0 ≤ a_ij ≤ 10⁶).

Output

Print a single number — the answer to the problem.

Sample Input

4 5
abcde
abcde
abcde
abcde
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1

Sample Output

HINT

题意

给你一堆字符串，如果这个字符的第i个字符在所有字符串的第i个都是不同的，那么我们就说这个字符串很好记

然后你可以花费 a[i][j]使得这个字符变成任意字符

然后问你最小花费，使得所有字符串都很好记

题解：

代码：

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 200001

#define mod 10007

#define eps 1e-9

int Num;

char CH[];

//const int inf=0x7fffffff;   //нчоч╢С

const int inf=<<;

/*

inline void P(int x)

{

    Num=0;if(!x){putchar('0');puts("");return;}

    while(x>0)CH[++Num]=x%10,x/=10;

    while(Num)putchar(CH[Num--]+48);

    puts("");

}

*/

inline ll read()

{

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

inline void P(int x)

{

    Num=;if(!x){putchar('');puts("");return;}

    while(x>)CH[++Num]=x%,x/=;

    while(Num)putchar(CH[Num--]+);

    puts("");

}

//**************************************************************************************

int a[][];

string s[];

int d[<<];

int p[];

int main()

{

    int n=read(),m=read();

    for(int i=;i<n;i++)

        cin>>s[i];

    for(int i=;i<n;i++)

        for(int j=;j<m;j++)

            cin>>a[i][j];

    for(int j=;j<<<n;j++)

        d[j]=inf;

    d[]=;

    for(int i=;i<m;i++)

    {

        for(int j=;j<;j++)

            p[j]=;

        for(int j=;j<n;j++)

            p[s[j][i]-'a']|=<<j;

        for(int j=;j<;j++)

        {

            int t=,mx=;

            for(int k=;k<n;k++)

            {

                if(p[j]>>k&)

                {

                    for(int l=;l<<<n;l++)

                        d[l|(<<k)]=min(d[l|(<<k)],d[l]+a[k][i]);

                    t+=a[k][i];

                    mx=max(mx,a[k][i]);

                }

            }

            for(int l=;l<<<n;l++)

                d[l|p[j]]=min(d[l|p[j]],d[l]+t-mx);

        }

    }

    cout<<d[(<<n)-]<<endl;

    return ;

}

巴特西

Codeforces Round #302 (Div. 1) C. Remembering Strings DP