BZOJ3864: Hero meet devil【dp of dp】

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

---

給你一個序列s，問長度爲m的和s最長公共子序列是\([0,|s|]\)的串有多少個

思路

dp of dp

由於最長公共子序列的dp矩陣每一行相鄰兩項不會超過1，因此就能夠用二進制狀壓

而後就先dp出每一個狀態加上一個字符的轉移

而後上狀壓dp就能夠了

注意dp數組的清零問題

---

#include<bits/stdc++.h>

using namespace std;

const int Mod = 1e9 + 7;
const int N = 16;
const int M = 1 << N;

int trans[M][4], bitcnt[M], dp[2][M];
int len, m, ans[N], f[N], g[N];
char s[N];

int add(int a, int b) {
  return (a += b) >= Mod ? a - Mod : a;
}

void solve() {
  scanf("%s%d", s + 1, &m);
  len = strlen(s + 1);
  for (int i = 1; i <= len; i++) {
    if (s[i] == 'A') s[i] = 0;
    if (s[i] == 'C') s[i] = 1;
    if (s[i] == 'G') s[i] = 2;
    if (s[i] == 'T') s[i] = 3;
  }
  for (int i = 0; i < (1 << len); i++) {
    for (int j = 1; j <= len; j++) {
      f[j] = f[j - 1] + ((i >> (j - 1)) & 1);
    }
    for (int j = 0; j < 4; j++) {
      for (int k = 1; k <= len; k++) {
        g[k] = max(g[k - 1], f[k]);
        if (s[k] == j) g[k] = max(g[k], f[k - 1] + 1);
      }
      trans[i][j] = 0;
      for (int k = 1; k <= len; k++) {
        trans[i][j] |= (g[k] - g[k - 1]) << (k - 1);
      }
    }
  }
  int ind = 0;
  for (int i = 0; i < (1 << len); i++) dp[ind][i] = 0; // **
  dp[ind][0] = 1;
  for (int i = 1; i <= m; i++) {
    ind ^= 1;
    for (int j = 0; j < (1 << len); j++) {
      dp[ind][j] = 0;
    }
    for (int j = 0; j < (1 << len); j++) if (dp[ind ^ 1][j]) {
      for (int k = 0; k < 4; k++) {
        dp[ind][trans[j][k]] = add(dp[ind][trans[j][k]], dp[ind ^ 1][j]);
      }
    }
  }
  for (int i = 0; i <= len; i++) ans[i] = 0;
  for (int i = 0; i < (1 << len); i++) {
    ans[bitcnt[i]] = add(ans[bitcnt[i]], dp[ind][i]); 
  }
  for (int i = 0; i <= len; i++) {
    printf("%d\n", ans[i]);
  }
}

int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif 
  for (int i = 0; i < M; i++) {
    for (int j = 1; j <= N; j++) {
      bitcnt[i] += (i >> (j - 1)) & 1;
    } 
  }
  int T; scanf("%d", &T);
  while (T--) solve();
  return 0;
}