（HDU）1298 （POJ）1451 - T9 【字典树】+【DFS】

T9

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2852    Accepted Submission(s): 1068

Problem Description

A while ago it was quite cumbersome to create a message for the Short Message Service (SMS) on a mobile phone. This was because you only have nine keys and the alphabet has more than nine letters, so most characters could only be entered by pressing one key several times. For example, if you wanted to type "hello" you had to press key 4 twice, key 3 twice, key 5 three times, again key 5 three times, and finally key 6 three times. This procedure is very tedious and keeps many people from using the Short Message Service.

This led manufacturers of mobile phones to try and find an easier way to enter text on a mobile phone. The solution they developed is called T9 text input. The "9" in the name means that you can enter almost arbitrary words with just nine keys and without pressing them more than once per character. The idea of the solution is that you simply start typing the keys without repetition, and the software uses a built-in dictionary to look for the "most probable" word matching the input. For example, to enter "hello" you simply press keys 4, 3, 5, 5, and 6 once. Of course, this could also be the input for the word "gdjjm", but since this is no sensible English word, it can safely be ignored. By ruling out all other "improbable" solutions and only taking proper English words into account, this method can speed up writing of short messages considerably. Of course, if the word is not in the dictionary (like a name) then it has to be typed in manually using key repetition again.


Figure 8: The Number-keys of a mobile phone.

More precisely, with every character typed, the phone will show the most probable combination of characters it has found up to that point. Let us assume that the phone knows about the words "idea" and "hello", with "idea" occurring more often. Pressing the keys 4, 3, 5, 5, and 6, one after the other, the phone offers you "i", "id", then switches to "hel", "hell", and finally shows "hello".

Write an implementation of the T9 text input which offers the most probable character combination after every keystroke. The probability of a character combination is defined to be the sum of the probabilities of all words in the dictionary that begin with this character combination. For example, if the dictionary contains three words "hell", "hello", and "hellfire", the probability of the character combination "hell" is the sum of the probabilities of these words. If some combinations have the same probability, your program is to select the first one in alphabetic order. The user should also be able to type the beginning of words. For example, if the word "hello" is in the dictionary, the user can also enter the word "he" by pressing the keys 4 and 3 even if this word is not listed in the dictionary.

 

Input

The first line contains the number of scenarios.

Each scenario begins with a line containing the number w of distinct words in the dictionary (0<=w<=1000). These words are given in the next w lines. (They are not guaranteed in ascending alphabetic order, although it's a dictionary.) Every line starts with the word which is a sequence of lowercase letters from the alphabet without whitespace, followed by a space and an integer p, 1<=p<=100, representing the probability of that word. No word will contain more than 100 letters.

Following the dictionary, there is a line containing a single integer m. Next follow m lines, each consisting of a sequence of at most 100 decimal digits 2-9, followed by a single 1 meaning "next word".

 

Output

The output for each scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1.

For every number sequence s of the scenario, print one line for every keystroke stored in s, except for the 1 at the end. In this line, print the most probable word prefix defined by the probabilities in the dictionary and the T9 selection rules explained above. Whenever none of the words in the dictionary match the given number sequence, print "MANUALLY" instead of a prefix.

Terminate the output for every number sequence with a blank line, and print an additional blank line at the end of every scenario.

 

Sample Input

2
5
hell 3
hello 4
idea 8
next 8
super 3
2
435561
43321
7
another 5
contest 6
follow 3
give 13
integer 6
new 14
program 4
5
77647261
6391
4681
26684371
77771

 

Sample Output

Scenario #1:
i
id
hel
hell
hello

i
id
ide
idea


Scenario #2:
p
pr
pro
prog
progr
progra
program

n
ne
new

g
in
int

c
co
con
cont
anoth
anothe
another

p
pr
MANUALLY
MANUALLY

 

题意：模拟手机输入法的记忆性，你以前输入过的单词会优先出现。（9宫格输入）

输入：第一行输入n，表示有n组数据。每组数据第一行为一个数w，在接下来w行，每行有一个字符串s和数字p（空格隔开），表示该字符串的输入次数。接下来一行为数字m，在接下来m行，每行是一串数字，表示模拟手机在9宫格输入法的按键。（1表示结束）

输出：问每次按键可能出现的字符串,按使用频率高的输出,若没有则输出MANUALLY。

分析：这题用字典树将字符串插入后，如何使得按键满足出现过的字符串呢？我们考虑第一个样例串，按下前两个数字的时候依次是i、id，按下第三个数字的时候，id前缀的字符串已经无法继续匹配了。那么程序需要回过头去需找满足输入的字符串，由于先后顺序，i在h的前面，所以之前没有显示h、he，而是显示i、id，直到id前缀串无法再匹配，出现的是hel、hell。这里面就体现了DFS（深度优先搜索）的思想，所以我们的代码需要用这样的方式来写。

这题蛮坑爹的，输出格式需要注意。以及储存九宫格键位的小技巧，在dfs函数里面体现到位。

我用的是数组字典树

#include <bits/stdc++.h>

using namespace std;

int key[10]={0,0,3,3,3,3,3,4,3,4};    //表示按键i有几个字符
int tot,n,w,p,m,kase=0;               //变量命名按照题述
char s[110];
string out;
int outLen;                           //这个变量单独为out字符串服务
char ch[10][5]={"","","abc","def",
             "ghi", "jkl","mno",
             "pqrs","tuv","wxyz"};    // 九宫格

struct node
{
    int next[26];
    int cnt;
    void init()
    {
        cnt=0;
        memset(next,-1,sizeof(next));
    }
}T[1000000];

void insert(char* s,int n)
{
    int i,p=0,len=strlen(s);
    for(i=0;i<len;i++)
    {
        int x=s[i]-'a';
        if(T[p].next[x]==-1)
        {
            T[tot].init();
            T[p].next[x]=tot++;
        }
        p=T[p].next[x];
        T[p].cnt+=n;
    }
}

void dfs(int i,int len,int p,string a)
{
    if(i==len && T[p].cnt>outLen)   //dfs终止条件，a字符串达到要求长度
    {
        outLen=T[p].cnt;
        out=a;
    }
    int x=s[i]-'0';                 //这里的x对应的是key字典
    for(int j=0;j<key[x];j++)
    {
        int y=ch[x][j]-'a';
        if(T[p].next[y]!=-1)            //如果有出现过的字母
            dfs(i+1,len,T[p].next[y],a+ch[x][j]);
    }
}

void search(char* s)
{
    int i,len=strlen(s);
    for(i=1;i<len;i++)       //按键for循环，按下i个键，出现长度为i的字符
    {
        outLen=0;
        dfs(0,i,0,"\0");
        if(outLen) cout<<out<<endl;
        else printf("MANUALLY\n");
    }
    printf("\n");
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("test.txt","r",stdin);
    freopen("test.out","w",stdout);
    #endif // ONLINE_JUDGE

    for(scanf("%d",&n);n;n--)
    {
        printf("Scenario #%d:\n",++kase);
        T[0].init(); tot=1;    //初始化字典树，类似于清空内存
        scanf("%d",&w);
        for(int i=0;i<w;i++)
        {
            scanf("%s %d",s,&p);
            insert(s,p);
        }
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            scanf("%s",s);
            search(s);
        }
        printf("\n");
    }
    return 0;
}

指针字典树+DFS写法

#include<iostream>  
#include<cstdio>  
#include<cstdlib>  
#include<cstring>  
#include<string>  
#include<queue>  
#include<algorithm>  
#include<map>  
#include<iomanip>  
#define INF 99999999  
using namespace std;  
  
const int MAX=100+10;  
int m[10]={0,0,3,3,3,3,3,4,3,4};//表示按键i有几个字符  
char ch[10][5]={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};   
char s[MAX];  
string temp;  
int sum;  
  
struct TrieNode{  
    int num;//出现频率  
    TrieNode *next[26];  
    TrieNode(){  
        num=0;  
        memset(next,0,sizeof next);  
    }   
}root;  
  
void InsertNode(char *a,int num){  
    int k=0;  
    TrieNode *p=&root;  
    while(a[k]){  
        if(!p->next[a[k]-'a'])p->next[a[k]-'a']=new TrieNode;  
        p=p->next[a[k++]-'a'];  
        p->num+=num;  
    }  
}  
  
void SearchTrie(int k,int len,TrieNode *p,string a){  
    if(k == len){  
        if(p->num > sum){  
            sum=p->num;  
            temp=a;  
        }  
    }  
    int t=s[k]-'0';  
    for(int i=0;i<m[t];++i){  
        if(p->next[ch[t][i]-'a'])SearchTrie(k+1,len,p->next[ch[t][i]-'a'],a+ch[t][i]);  
    }  
}  
  
void Free(TrieNode *p){  
    for(int i=0;i<26;++i)if(p->next[i])Free(p->next[i]);  
    delete p;  
}  
  
int main(){  
    int t,n,num=0,w;  
    cin>>t;  
    while(t--){  
        cout<<"Scenario #"<<++num<<":\n";  
        cin>>n;  
        for(int i=0;i<n;++i){  
            cin>>s>>w;  
            InsertNode(s,w);  
        }  
        cin>>n;  
        for(int i=0;i<n;++i){  
            cin>>s;  
            int len=strlen(s);  
            for(int j=1;j<len;++j){  
                sum=0;  
                SearchTrie(0,j,&root,"\0");  
                if(sum)cout<<temp<<endl;  
                else cout<<"MANUALLY"<<endl;  
            }  
            cout<<endl;  
        }  
        cout<<endl;  
        for(int i=0;i<26;++i){  
            if(root.next[i])Free(root.next[i]);  
            root.next[i]=NULL;  
        }   
    }  
    return 0;  
}