NBUT 1222 English Game（trie树+DP）

• [1222] English Game

• 时间限制: 1000 ms　内存限制: 131072 K
• 问题描述

• 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.

  
  

  
  

  
  

  
  

  
  

  
  

  
  

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

• 1 aaaaa 23 aaaa 2aa 5aaa 64 abca 1bc 2ab 4c 13 abcdab 10bc 20cd 303 abcdcd 100abc 1000bcd 10000

• 样例输出

• 87540-1

题目大意：

给出一个目标字符串和n个字符串及其对应的权值，求用这些字符串中的1个或多个组成目标字符串的最大权值。

解题思路：
用trie树保存b个字符串及其权值，接下来就是动态规划了。
到达某一点i，遍历trie树找到i+1为第一个字符的字符串，在结束位置j，dp[j]=max(dp[j],dp[i]+val[i+1到j之间的字符串])。除第0位之外，如果dp[i]=0说明没有字符串能够组成0到i之间的字符串，那么就不用查询。

参考代码：

#include<map>#include<stack>#include<queue>#include<cmath>#include<vector>#include<cctype>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const double eps=1e-10;const int INF=0x3f3f3f3f;const int MAXN=1e4+50;typedef long long LL;struct node{    node* next[26];    int val;    node()    {        val=0;        memset(next,0,sizeof(next));    }}*root;int n,L,dp[MAXN];char s[MAXN];void trie_insert(node* start,char* word,int v){    node* now=start;    int l=strlen(word);    for(int i=0; i<l; i++)    {        int id=word[i]-'a';        if(now->next[id]==NULL)            now->next[id]=new node();        now=now->next[id];    }    now->val=v;}int trie_query(node* start,int pos){    node* now=start;    int i;    for(i=pos+1; i<=L; i++)    {        int id=s[i]-'a';        if(now->next[id]==NULL)            break;        now=now->next[id];        if(now->val)            dp[i]=max(dp[i],dp[pos]+now->val);    }}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);#endif // ONLINE_JUDGE    while(scanf("%d%s",&n,s+1)!=EOF)    {        root=new node();        L=strlen(s+1);        memset(dp,0,sizeof(dp));        for(int i=0; i<n; i++)        {            char tmp[MAXN];            int d;            scanf("%s%d",tmp,&d);            trie_insert(root,tmp,d);        }        for(int i=0; i<=L; i++)            if(dp[i]||i==0)//如果dp[i]=0说明没有字符串能够组成0到i之间的字符串，那么就不用查询                trie_query(root,i);        printf("%d\n",dp[L]==0?-1:dp[L]);    }    return 0;}