HDU 3341 Lost's revenge AC自动机+dp

Lost's revenge

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3757    Accepted Submission(s): 1020

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

3

AC
CG
GT

CGAT

1

AA

AAA

0

 

Sample Output

Case 1: 3

Case 2: 2

/*
HDU 3341 Lost's revenge AC自动机+dp

给n个子串和一个字符串str,str中的位置可以随便调整.求最多可能包含多少个子串
很明显使用dp,但是在保存状态的时候出现问题. 因为有AGCT四个,那么最大需要的内存
就是40*40*40*40 超内存

但实际上是str的总长度为40,即 A+C+G+T的个数为40,那么最大需要内存的也就是10*10*10*10
这样的话dp[i][j]来保存 当前节点为i,AGCT的使用状态为j时的最大值;   //dp是硬伤TAT
然后就是重复的子串也要重复计算。

hhh-2016-04-27 22:11:06
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef unsigned long long ll;
typedef unsigned int ul;
const int mod = 20090717;
const int INF = 0x3f3f3f3f;
const int N = 505;
int tot;
int n;
int dp[N][11*11*11*11+10];
int tal[10];
int num[4];
int can(int id,int wt)
{
    if(wt == 0)
        return id/1000000;
    else if(wt == 1)
        return id/10000%100;
    else if(wt == 2)
        return id/100%100;
    else if(wt == 3)
        return id%100;
}

struct Tire
{
    int nex[N][4],fail[N],ed[N];
    int root,L;
    int newnode()
    {
        for(int i = 0; i < 4; i++)
            nex[L][i] = -1;
        ed[L++] = 0;
        return L-1;
    }

    void ini()
    {
        L = 0,root = newnode();
    }

    int cal(char ch)
    {
        if(ch == 'A')
            return 0;
        else if(ch == 'C')
            return 1;
        else if(ch == 'G')
            return 2;
        else if(ch == 'T')
            return 3;
    }

    void inser(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0; i < len; i++)
        {
            int ta = cal(buf[i]);
            if(nex[now][ta] == -1)
                nex[now][ta] = newnode();
            now = nex[now][ta];
        }
        ed[now]++;
    }

    void build()
    {
        queue<int >q;
        fail[root] = root;
        for(int i = 0; i < 4; i++)
            if(nex[root][i] == -1)
                nex[root][i] = root;
            else
            {
                fail[nex[root][i]] = root;
                q.push(nex[root][i]);
            }
        while(!q.empty())
        {
            int now = q.front();
            q.pop();
            ed[now] += ed[fail[now]];
            for(int i = 0; i < 4; i++)
            {
                if(nex[now][i] == -1)
                    nex[now][i] = nex[fail[now]][i];
                else
                {
                    fail[nex[now][i]] = nex[fail[now]][i];
                    q.push(nex[now][i]);
                }
            }
        }
    }

    void solve(int len)
    {
        memset(dp,-1,sizeof(dp));
        dp[0][0] = 0;
        num[0] = (tal[3]+1)*(tal[1]+1)*(1+tal[2]);
        num[1] = (tal[2]+1)*(tal[3]+1);
        num[2] = tal[3]+1;
        num[3] = 1;
        for(int t0 = 0; t0 <= tal[0]; t0++)
            for(int t1 = 0; t1 <= tal[1]; t1++)
                for(int t2 = 0; t2 <= tal[2]; t2++)
                    for(int t3 = 0; t3 <= tal[3]; t3++)
                        for(int i = 0; i < L; i++)
                        {
                            int tn = t1*num[1]+t2*num[2]+t3+t0*num[0];
                            if(dp[i][tn] >= 0)
                                for(int k = 0; k < 4; k++)
                                {
                                    int ta = nex[i][k];
                                    if(k == 0 && t0 == tal[0])continue;
                                    if(k == 1 && t1 == tal[1])continue;
                                    if(k == 2 && t2 == tal[2])continue;
                                    if(k == 3 && t3 == tal[3])continue;
                                    dp[ta][tn+num[k]] = max(dp[ta][tn+num[k]],dp[i][tn] + ed[ta]);
                                }
                        }

        int ans = 0;
        int ta = num[0]*tal[0]+num[1]*tal[1]+num[2]*tal[2]+num[3]*tal[3];
        for(int i =0 ; i < L; i++)
        {
            ans = max(ans,dp[i][ta]);
        }
        printf("%d\n",ans);
    }
};

Tire ac;
char buf[50];

int main()
{
    int cas = 1;
    while(scanf("%d",&n)==1 && n)
    {
        ac.ini();

        for(int i = 0; i < n; i++)
        {
            scanf("%s",buf);
            ac.inser(buf);
        }
        ac.build();
        scanf("%s",buf);
        memset(tal,0,sizeof(tal));
        for(int i = 0; i < (int)strlen(buf) ; i++)
        {
            tal[ac.cal(buf[i])]++;
        }
        printf("Case %d: ",cas++);
        ac.solve(strlen(buf));
    }
    return 0;
}