HDU 2296 Ring （AC自动机+DP）

Ring

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1564    Accepted Submission(s): 487

Problem Description

For the hope of a forever love, Steven is planning to send a ring to Jane with a romantic string engraved on. The string's length should not exceed N. The careful Steven knows Jane so deeply that he knows her favorite words, such as "love", "forever". Also, he knows the value of each word. The higher value a word has the more joy Jane will get when see it.
The weight of a word is defined as its appeared times in the romantic string multiply by its value, while the weight of the romantic string is defined as the sum of all words' weight. You should output the string making its weight maximal. 

 

 

Input

The input consists of several test cases. The first line of input consists of an integer T, indicating the number of test cases. Each test case starts with a line consisting of two integers: N, M, indicating the string's length and the number of Jane's favorite words. Each of the following M lines consists of a favorite word Si. The last line of each test case consists of M integers, while the i-th number indicates the value of Si.
Technical Specification

1. T ≤ 15
2. 0 < N ≤ 50, 0 < M ≤ 100.
3. The length of each word is less than 11 and bigger than 0.
4. 1 ≤ Hi ≤ 100. 
5. All the words in the input are different.
6. All the words just consist of 'a' - 'z'.

 

 

Output

For each test case, output the string to engrave on a single line.
If there's more than one possible answer, first output the shortest one. If there are still multiple solutions, output the smallest in lexicographically order.

The answer may be an empty string. 

 

 

Sample Input

2 7 2 love ever 5 5 5 1 ab 5

 

 

Sample Output

lovever abab

Hint

Sample 1: weight(love) = 5, weight(ever) = 5, so weight(lovever) = 5 + 5 = 10 Sample 2: weight(ab) = 2 * 5 = 10, so weight(abab) = 10

 

 

Source

The 4th Baidu Cup final

 

 

Recommend

lcy

 

 

 

 

 

比较麻烦的是需要输出字典序最小的解。

 

增加个字符串来记录

 

//============================================================================

// 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;



int a[110];

int dp[55][1110];

char str[55][1110][55];



bool cmp(char s1[],char s2[])

{

    int len1=strlen(s1);

    int len2=strlen(s2);

    if(len1 != len2)return len1 < len2;

    else return strcmp(s1,s2) < 0;

}



const int INF=0x3f3f3f3f;

struct Trie

{

    int next[1110][26],fail[1110],end[1110];

    int root,L;

    int newnode()

    {

        for(int i = 0;i < 26;i++)

            next[L][i] = -1;

        end[L++] = -1;

        return L-1;

    }

    void init()

    {

        L = 0;

        root = newnode();

    }

    void insert(char buf[],int id)

    {

        int len = strlen(buf);

        int now = root;

        for(int i = 0;i < len;i++)

        {

            if(next[now][buf[i]-'a'] == -1)

                next[now][buf[i]-'a'] = newnode();

            now = next[now][buf[i]-'a'];

        }

        end[now] = id;

    }

    void build()

    {

        queue<int>Q;

        fail[root] = root;

        for(int i = 0;i < 26;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();

            for(int i = 0;i < 26;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]);

                }

        }

    }

    void solve(int n)

    {

        for(int i = 0;i <= n;i++)

            for(int j = 0;j < L;j++)

                dp[i][j] = -INF;

        dp[0][0] = 0;

        strcpy(str[0][0],"");

        char ans[55];

        strcpy(ans,"");

        int Max = 0;

        char tmp[55];

        for(int i = 0; i < n;i++)

            for(int j = 0;j < L;j++)

                if(dp[i][j]>=0)

                {

                    strcpy(tmp,str[i][j]);

                    int len = strlen(tmp);

                    for(int k = 0;k < 26;k++)

                    {

                        int nxt=next[j][k];

                        tmp[len] = 'a'+k;

                        tmp[len+1] = 0;

                        int tt = dp[i][j];

                        if(end[nxt] != -1)

                            tt+=a[end[nxt]];



                        if(dp[i+1][nxt]<tt || (dp[i+1][nxt]==tt && cmp(tmp,str[i+1][nxt])))

                        {

                            dp[i+1][nxt] = tt;

                            strcpy(str[i+1][nxt],tmp);

                            if(tt > Max ||(tt==Max && cmp(tmp,ans)))

                            {

                                Max = tt;

                                strcpy(ans,tmp);

                            }

                        }

                    }

                }

        printf("%s\n",ans);

    }

};

char buf[60];

Trie ac;

int main()

{

//    freopen("in.txt","r",stdin);

//    freopen("out.txt","w",stdout);

    int T;

    int n,m;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d%d",&n,&m);

        ac.init();

        for(int i = 0;i < m;i++)

        {

            scanf("%s",buf);

            ac.insert(buf,i);

        }

        for(int i = 0;i < m;i++)

            scanf("%d",&a[i]);

        ac.build();

        ac.solve(n);

    }

    return 0;

}