BZOJ3864: Hero meet devil(dp套dp)

Time Limit: 8 Sec Memory Limit: 128 MB
Submit: 397 Solved: 206
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Description

There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli.

After the ring has been destroyed, the devil doesn't feel angry, and she is attracted by z*p's wisdom and handsomeness. So she wants to find z*p out.

But what she only knows is one part of z*p's DNA sequence S leaving on the broken ring.

Let us denote one man's DNA sequence as a string consist of letters from ACGT. The similarity of two string S and T is the maximum common subsequence of them, denote by LCS(S,T).

After some days, the devil finds that. The kingdom's people's DNA sequence is pairwise different, and each is of length m. And there are 4^m people in the kingdom.

Then the devil wants to know, for each 0 <= i <= |S|, how many people in this kingdom having DNA sequence T such that LCS(S,T) = i.

You only to tell her the result modulo 10^9+7.

Input

The first line contains an integer T, denoting the number of the test cases.

For each test case, the first line contains a string S. the second line contains an integer m.

T<=5

|S|<=15. m<= 1000.

Output

For each case, output the results for i=0,1,...,|S|, each on a single line.

Sample Input

1
GTC
10

Sample Output

1
22783
528340
497452

HINT

Source

首先想一下LCS的转移方程

$$lcs[i][j]=max \begin{cases} lcs[i-1][j-1]+1 & \text{if t[i]=s[j]} \\ lcs[i-1][j] \\ lcs[i][j-1] \end{cases}$$

这样的话，当$i$确定是，$lcs[i][j]$和$lcs[i][j-1]$最多相差$1$

且题目中说$|S|<= 15$，因此我们考虑把差分后的lcs数组状压起来

那么如何统计答案呢？

设$f[i][sta]$表示在第$i$个位置，此时lcs的状态为$sta$的方案数，

然后我们枚举一下这个位置选ACGT中的哪个

设$trans[sta'][A/C/G/T]$为在$sta$状态表示的lcs后加了ACGT中的一个后的状态，这个很显然可以预处理得到

那么转移方程为

$$f[i][ trans[sta][k] ] += f[i - 1][sta] $$

$$f[0][0] = 1$$

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

const int  MAXN = , mod = 1e9 + ;

char S[], SS[] = {"ACGT"};

int a[], f[MAXN][( << ) + ], trans[( << ) + ][], N, Len, limit, ans[];

int tmp[][];

int solve(int sta, int ch) {

    int ret = ;

    memset(tmp, , sizeof(tmp));

    for(int i = ; i < N; i++) tmp[][i + ] = tmp[][i] + ((sta >> i) &  );

    for(int i = ; i <= N; i++) {

        int mx = ;

        if(a[i] == ch) mx = tmp[][i - ] + ;

        mx = max( max(mx, tmp[][i]), tmp[][i-]);

        tmp[][i] = mx;

    }

    for(int i = ; i < N; i++) ret += ( << i) * (tmp[][i + ] - tmp[][i]);

    return ret;

}

int main() {

    #ifdef WIN32

    freopen("a.in", "r", stdin);

    #endif

    int QWQ;scanf("%d", &QWQ);

    while(QWQ--) {

        memset(f, , sizeof(f));memset(ans, , sizeof(ans));

        scanf("%s", S + );

        N = strlen(S + ); limit = ( << N) - ;

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

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

                if(S[i] == SS[j]){a[i] = j + ;break;}

        scanf("%d", &Len);

        f[][] = ;

        for(int sta = ; sta <= limit; sta++)

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

                trans[sta][j] = solve(sta, j);

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

            for(int sta = ; sta <= limit; sta++)

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

                    f[i][ trans[sta][k] ] = (f[i][ trans[sta][k] ] + f[i - ][sta]) % mod;

        for(int sta = ; sta <= limit; sta++)

            ans[__builtin_popcount(sta)] = (ans[__builtin_popcount(sta)] + f[Len][sta]) % mod;

            //这个函数是算出sta中1的个数

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

            printf("%d\n", ans[i] % mod);

    }

    return ;

}