hdu 3341 Lost's revenge



Lost's revenge

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 4207    Accepted Submission(s): 1166

Problem Description

Lost and AekdyCoin are friends. They always play "number game"(A boring game based on number theory) together. We all know that AekdyCoin is the man called "nuclear weapon of FZU,descendant of Jingrun", because of his talent in the field of number theory. So Lost had never won the game. He was so ashamed and angry, but he didn't know how to improve his level of number theory.

One noon, when Lost was lying on the bed, the Spring Brother poster on the wall(Lost is a believer of Spring Brother) said hello to him! Spring Brother said, "I'm Spring Brother, and I saw AekdyCoin shames you again and again. I can't bear my believers were being bullied. Now, I give you a chance to rearrange your gene sequences to defeat AekdyCoin!".

It's soooo crazy and unbelievable to rearrange the gene sequences, but Lost has no choice. He knows some genes called "number theory gene" will affect one "level of number theory". And two of the same kind of gene in different position in the gene sequences will affect two "level of number theory", even though they overlap each other. There is nothing but revenge in his mind. So he needs you help to calculate the most "level of number theory" after rearrangement.

 

Input

There are less than 30 testcases.
For each testcase, first line is number of "number theory gene" N(1<=N<=50). N=0 denotes the end of the input file.
Next N lines means the "number theory gene", and the length of every "number theory gene" is no more than 10.
The last line is Lost's gene sequences, its length is also less or equal 40.
All genes and gene sequences are only contains capital letter ACGT.

 

Output

For each testcase, output the case number(start with 1) and the most "level of number theory" with format like the sample output.

 

Sample Input

3ACCGGTCGAT1AAAAA0

 

Sample Output

Case 1: 3Case 2: 2

 

Author

Qinz@XDU

 

Source

HDOJ Monthly Contest – 2010.03.06

 

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【分析】

AC自动机+dp

首先把所有小串建一个AC自动机

然后用dp[j][a][b][c][d]表示走到自动机的节点 j，而且现在用了文本串中的a个A，b个G，c个C，d个T所能匹配到的最多的小串数目。

发现空间炸了...

观察到文本串长度<=40，那么最多的组合应该是有10*10*10*10=1W种方式组合，那么我们把a,b,c,d不要开四维空间了，hash一下变成一维来防止MLE。

那么方程很好写了，具体请看代码实现吧。

还有就是要注意循环顺序，节点应该在最后面的循环。（WA哭了QAQ）

【代码】

//hdu 3341 Lost's revenge#include<iostream>#include<cstring>#include<cstdio>#include<queue>#define ll long long#define M(a) memset(a,0,sizeof a)#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;const int mxn=505;char s[mxn];queue <int> q;int n,m,T,num;int has[41][41][41][41];int map[150],tot[5],dp[mxn][11*11*11*11+5];struct node {int son[5],fail,cnt;} a[mxn];inline void clear(int x){    a[x].fail=a[x].cnt=0;    memset(a[x].son,0,sizeof a[x].son);}inline void trie(){    scanf("%s",s+1);    int i,j,x=0,len=strlen(s+1);    fo(i,1,len)    {        int c=map[s[i]];        if(!a[x].son[c])          clear(++num),a[x].son[c]=num;        x=a[x].son[c];    }    a[x].cnt++;}inline void build(){    int i,j;    fo(i,1,4) if(a[0].son[i]) q.push(a[0].son[i]);    while(!q.empty())    {        int x=q.front();q.pop();        int fail=a[x].fail;        fo(i,1,4)        {            int y=a[x].son[i];            if(y) a[y].fail=a[fail].son[i],q.push(y);            else a[x].son[i]=a[fail].son[i];        }        a[x].cnt+=a[fail].cnt;    }}inline void init(){    int A,B,C,D,czy=-1;    fo(A,0,tot[1]) fo(B,0,tot[2]) fo(C,0,tot[3]) fo(D,0,tot[4])       has[A][B][C][D]=++czy;}inline int dynamic(){    init();    int i,j,k,A=0,B=0,C=0,D=0,ans=0;    memset(dp,-0x3f,sizeof dp);    dp[0][0]=0;    fo(A,0,tot[1]) fo(B,0,tot[2]) fo(C,0,tot[3]) fo(D,0,tot[4]) fo(k,1,4) fo(j,0,num)     {        int S=has[A][B][C][D],s1,s2,s3,s4,tmp=a[a[j].son[k]].cnt;        if(D && k==4) s1=has[A][B][C][D-1],dp[a[j].son[k]][S]=max(dp[a[j].son[k]][S],dp[j][s1]+tmp);        if(C && k==3) s2=has[A][B][C-1][D],dp[a[j].son[k]][S]=max(dp[a[j].son[k]][S],dp[j][s2]+tmp);        if(B && k==2) s3=has[A][B-1][C][D],dp[a[j].son[k]][S]=max(dp[a[j].son[k]][S],dp[j][s3]+tmp);        if(A && k==1) s4=has[A-1][B][C][D],dp[a[j].son[k]][S]=max(dp[a[j].son[k]][S],dp[j][s4]+tmp);//      if(dp[a[j].son[k]][S]>=0)//        printf("树上节点%d 儿子 %d=%d %d %d %d=%d,new=%d\n",j,a[j].son[k],A,B,C,D,dp[a[j].son[k]][S],k);    }    int end=has[tot[1]][tot[2]][tot[3]][tot[4]];    fo(j,0,num) ans=max(ans,dp[j][end]);    return ans;}int main(){    int i,j;    map['A']=1,map['G']=2,map['C']=3,map['T']=4;    while(scanf("%d",&n) && n)    {        M(tot);        T++,num=0,clear(0);        fo(i,1,n) trie();        build();        scanf("%s",s+1);        int len=strlen(s+1);        fo(i,1,len) tot[map[s[i]]]++;        printf("Case %d: %d\n",T,dynamic());    }    return 0;}/*3GTGGAGAGGT*/