Hero meet devil

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 692    Accepted Submission(s): 338 Problem 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   Author WJMZBMR   Source 2014 Multi-University Training Contest 4 

题目大意：

    给你一个DNA序列，求有多少个长度为N的DNA序列和给定序列的LCS为0.1.2....

解题思路：

    首先比较容易想到使用dp[i][s]表示：当前长度为i，给出序列中已经匹配的位置的下标状压为s，方案数。

    不过我们可以发现，如果按照普通求LCS的方式进行转移的话，我们需要枚举当前添加一个字符可以转移到的所有状态再取最大值，但我们这是计数DP，这样写的话，显然会重复计算。所以我们可以考虑再套一层DP，先预处理求出任意一个状态添加一个字符可以转移到的状态，然后计数DP时直接根据这个结果向最优的方向转移即可。

AC代码：

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <vector> #include <queue> using namespace std; #define mem(a,b) memset((a),(b),sizeof(a)) const int MOD=1000000007; const int MAXL=15+1; const int MAXS=1<<MAXL; const char d[]="ATCG"; char dna[MAXL]; int N, len, dp[2][MAXS], ans[MAXL]; int next_s[MAXS][4];//当前状态下，选择第j个字符，下一个状态 int pre[MAXL];//当前状态前i位有多少个1 int lcs[MAXL]; void init() { mem(dp, 0); mem(ans, 0); } inline void mod_add(int &a, const int &b) { a+=b; if(a>=MOD) a-=MOD; } void pre_work()//预处理出，每一个状态添加一个字符后，转移到的新状态 { for(int s=0;s<(1<<len);++s) { pre[0]=0; for(int i=1;i<=len;++i) pre[i]=pre[i-1]+((s>>(i-1))&1); for(int k=0;k<4;++k) { for(int i=1;i<=len;++i) { //通过在任意位置添加一个这个字符，前i位最大可以得到的LCS if(dna[i]==d[k]) lcs[i]=pre[i-1]+1; else lcs[i]=max(lcs[i-1], pre[i]); } //通过比较相邻两位的lcs，得到新状态的状压表示 int &t=next_s[s][k]=0; for(int i=1;i<=len;++i) t|=((lcs[i]!=lcs[i-1])<<(i-1)); } } } int main() { int T_T; scanf("%d", &T_T); while(T_T--) { scanf("%s%d", dna+1, &N); len=strlen(dna+1); init(); pre_work(); bool now=0, next=1; dp[now][0]=1; for(int i=0;i<N;++i) { for(int st=0;st<(1<<len);++st) for(int k=0;k<4;++k) mod_add(dp[next][next_s[st][k]], dp[now][st]); mem(dp[now], 0); now^=true; next^=true; } for(int i=0;i<(1<<len);++i) mod_add(ans[__builtin_popcount(i)], dp[now][i]); for(int i=0;i<=len;++i) printf("%d\n", ans[i]); } return 0; }