NEFU English Game 字符串 dp 字典树

English Game

Time Limit 1000ms

Memory Limit 65536K

description

　　This English game is a simple English words connection game.
　　The rules are as follows: there are N English words in a dictionary, and every word has its own weight v. There is a weight if the corresponding word is used. Now there is a target string X. You have to pick some words in the dictionary, and then connect them to form X. At the same time, the sum weight of the words you picked must be the biggest.
							

input

　　There are several test cases. For each test, N (1<=n<=1000) and X (the length of x is not bigger than 10000) are given at first. Then N rows follow. Each row contains a word wi (the length is not bigger than 30) and the weight of it. Every word is composed of lowercases. No two words in the dictionary are the same.  
							

output

　　For each test case, output the biggest sum weight, if you could not form the string X, output -1.
							

sample_input

　　1 aaaa
　　a 2
　　3 aaa
　　a 2
　　aa 5
　　aaa 6
　　4 abc
　　a 1
　　bc 2
　　ab 4
　　c 1
　　3 abcd
　　ab 10
　　bc 20
　　cd 30
　　3 abcd
　　cd 100
　　abc 1000
bcd 10000
							

sample_output

　　8
　　7
　　5
　　40
-1
							

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f[i]=max(f[j]+cost) 字符串j+1...i存在

用字典树优化

---------------------

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>

using namespace std;

char str[11111];
char words[1111][44];
int f[11111];
int _cost;
int n;
int pot;

//用类，或者结构体定义都行
struct trie
{
    int next[26];
    int value;
    int cost;
    trie()
    {
       for(int i=0;i<26;i++) next[i]=0;
       value=0;//记录是不是一个单词
       cost=0;
    }
    void init()
    {
        memset(next,0,sizeof(next));
        value=0;
        cost=0;
    }
}mmr[333333];

//插入：
void insert(int root,char* s,int ct)
{
    int p=root;
    int k=0;
    while(s[k]!='\0')
    {
        if(!mmr[p].next[s[k]-'a'])
        {
            mmr[pot].init();
            mmr[p].next[s[k]-'a']=pot++;
        }
        p=mmr[p].next[s[k]-'a'];
        k++;
    }
    mmr[p].value=1;
    mmr[p].cost=ct;
}

//查找
void find(int root,char* s,int pos)
{
    int p=root;
    int k=0;
    while(s[k]!='\0'&&mmr[p].next[s[k]-'a'])
    {
        p=mmr[p].next[s[k]-'a'];
        if (mmr[p].value==1)
        {
            f[pos+k+1]=max(f[pos+k+1],f[pos]+mmr[p].cost);
        }
        k++;
    }
}

int main()
{
    int l;
    int root;
    while (~scanf("%d%s",&n,str+1))
    {
        memset(f,-1,sizeof(f));
        root=1;
        mmr[root].init();
        pot=2;
        l=strlen(str+1);
        for (int i=0;i<n;i++)
        {
            scanf("%s",words[i]);
            scanf("%d",&_cost);
            insert(root,words[i],_cost);
        }
        f[0]=0;
        for (int i=1;i<=l;i++)
        {
            if (f[i-1]!=-1)
            {
                find(root,str+i,i-1);
            }
        }
        printf("%d\n",f[l]);
    }
    return 0;
}