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): 2262    Accepted Submission(s): 565


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
 

 

Author
Qinz@XDU
 

 

Source
 

 

Recommend
lcy
 
 
 
 
字符最长是40
只需要记录ACGT出现的次数。
 
如果使用5维数组,显然超内存了。
 
假设ACGT的总数分别为num[0],num[1],num[2],num[3]
 
那么对于ACGT的数量分别为ABCD的状态可以记录为:
 
A*(num[1]+1)*(num[2]+1)*(num[3]+1) + B*(num[2]+1)*(num[3]+1)+ C*(num[3]+1) +D
 
这样的状态最大就是11*11*11*11
 
复杂度也可以承受了。
 
 
字符串可能有重复的,用int来记录数量
 
 
//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;
const int INF = 0x3f3f3f3f;
struct Trie
{
    int next[510][4],fail[510];
    int end[510];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 4;i++)
            next[L][i] = -1;
        end[L++] = 0;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    int getch(char ch)
    {
        if(ch == 'A')return 0;
        else if(ch == 'C')return 1;
        else if(ch == 'G')return 2;
        else return 3;
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][getch(buf[i])] == -1)
                next[now][getch(buf[i])] = newnode();
            now = next[now][getch(buf[i])];
        }
        end[now] ++;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 4;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            end[now] += end[fail[now]];/********/
            for(int i = 0;i < 4;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int dp[510][11*11*11*11+5];
    int bit[4];
    int num[4];
    int solve(char buf[])
    {
        int len = strlen(buf);
        memset(num,0,sizeof(num));
        for(int i = 0;i < len;i++)
            num[getch(buf[i])]++;
        bit[0] = (num[1]+1)*(num[2]+1)*(num[3]+1);
        bit[1] = (num[2]+1)*(num[3]+1);
        bit[2] = (num[3]+1);
        bit[3] = 1;
        memset(dp,-1,sizeof(dp));
        dp[root][0] = 0;
        for(int A = 0;A <= num[0];A++)
            for(int B = 0;B <= num[1];B++)
                for(int C = 0;C <= num[2];C++)
                    for(int D = 0;D <= num[3];D++)
                    {
                        int s = A*bit[0] + B*bit[1] + C*bit[2] + D*bit[3];
                        for(int i = 0;i < L;i++)
                            if(dp[i][s] >= 0)
                            {
                                for(int k = 0;k < 4;k++)
                                {
                                    if(k == 0 && A == num[0])continue;
                                    if(k == 1 && B == num[1])continue;
                                    if(k == 2 && C == num[2])continue;
                                    if(k == 3 && D == num[3])continue;
                                    dp[next[i][k]][s+bit[k]] = max(dp[next[i][k]][s+bit[k]],dp[i][s]+end[next[i][k]]);
                                }
                            }
                    }
        int ans = 0;
        int status = num[0]*bit[0] + num[1]*bit[1] + num[2]*bit[2] + num[3]*bit[3];
        for(int i = 0;i < L;i++)
            ans = max(ans,dp[i][status]);
        return ans;
    }
};
char buf[50];
Trie ac;
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n;
    int iCase = 0;
    while( scanf("%d",&n) == 1 && n)
    {
        iCase++;
        ac.init();
        for(int i = 0;i < n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        scanf("%s",buf);
        printf("Case %d: %d\n",iCase,ac.solve(buf));
    }
    return 0;
}

 

 
 
 
 
 
 
 
 
 
 
 

posted on 2013-06-30 16:25  kuangbin  阅读(...)  评论(...编辑  收藏

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