HDU - 1298 T9(字典树+dfs)

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题意:给出一系列的单词作为字典,这些单词有着自己对应的出现概率对于有相同前缀的,前缀概率为包含的所有单词的概率相加,然后对应手机拼音九键的输入法规则且概率大的先输出,输入按下的键值,要求输出对应的系列字母,如果字典中不存在,则输出"MANUALLY".

分析:百度看了一下题解,发现还是用指针写的多,我还是想用数组实现.用字典树存储输入的单词,字典树的结构为child[10000][26]+value[10000].查询的时候用深搜TrieSearch(rt,cur,deep),具体看注释.

参考代码:

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
int w,m;
char word[105];//输入的单词
int pro;//输入单词对应值
char query[105];//按键
//TrieTree
int child[10000][26];//结点
int node;//记录结点的个数
int value[10000];//记录结点的值
//按键记录
int press[10] = {0,0,3,3,3,3,3,4,3,4};//每个键具有的字母个数
char key[10][5] ={"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};//每个键对应的字母
char ans[105];//记录查到的结果
char fi[105];//最终ans
int tmp;//查找过程中记录临时value

//初始化
void init()
{
mem(child[0],0);
node = 1;
mem(value,0);
}

//单词插入到树中
void TrieInsert( char word[], int pro)
{
int rt = 0;
int p = 0;
while( word[p] != '\0')
{
int id = word[p]-'a';
if( !child[rt][id])
{
mem(child[node],0);
value[node] = 0;
child[rt][id] = node++;
}
value[child[rt][id]] += pro;
rt = child[rt][id];
p++;
}
}

void TrieSearch( int rt, int cur, int deep)
{
if( cur == deep)
{
if( value[rt] > tmp)
{
tmp = value[rt];
for( int i = 0; i <= deep; i++)
fi[i] = ans[i];
fi[deep+1] = '\0';
}
return;
}

int len = press[query[cur+1]-'0'];//键包括的字母长度
for( int i = 0; i < len; i++)
{
int id = key[query[cur+1]-'0'][i]-'a';
if( child[rt][id])
{
ans[cur+1] = 'a'+id;
ans[cur+2]='\0';
TrieSearch(child[rt][id],cur+1,deep);
}
}
}

int main()
{
int T;
scanf("%d",&T);
while( T--)
{
init();
scanf("%d",&w);
for( int i = 1; i <= w; i++)
{
scanf("%s%d",word,&pro);
TrieInsert(word,pro);
}

static int cas = 1;
printf("Scenario #%d:\n",cas++);
scanf("%d",&m);
for( int i = 1; i <= m; i++)
{
scanf("%s",query);
for( int i = 0; query[i] != '1'; i++)
{
tmp = 0;
TrieSearch(0,-1,i);
if( tmp)
printf("%s\n",fi);
else
puts("MANUALLY");
}
puts("");
}
puts("");
}

return 0;
}