hdu4287(字典树)

来源：互联网发布：java 内存泄露几种情况编辑：程序博客网时间：2024/06/06 20:23

题目链接

字典树不会的同学:点这里

题意: 手机上有如下映射

2 -> a,b,c

...

9-> w x y z

给你n个数字和 m个单词 ,问每个数字对应的映射能组成多少种单词.依次输出.

一开始做的时候脑子进水了 ,把字母存入字典树里,结果就要考虑数字对应的字母情况.

比如 23对应 ad ae af bd be bf cd ce cf.这样内存不够.

我们把数字存入字典树,定义一个数组完成字母到数字的映射:

<span style="font-size:18px;">#include <stdio.h>#include <string.h>#include <stdlib.h>char word[5005][10];int ans[5005];int sign[129];typedef struct ac{int num;        //记录是第几个数struct ac *child[10];}tree;tree* root;void init(){        //初始化    sign['a']=sign['b']=sign['c']=2;    sign['d']=sign['e']=sign['f']=3;    sign['g']=sign['h']=sign['i']=4;    sign['j']=sign['k']=sign['l']=5;    sign['m']=sign['n']=sign['o']=6;    sign['p']=sign['q']=sign['r']=sign['s']=7;    sign['t']=sign['u']=sign['v']=8;    sign['w']=sign['x']=sign['y']=sign['z']=9;}void clear(tree * p) {     //清缓存 if(p==NULL) return ;elsefor(int i=0;i<=9;i++) clear(p->child[i]);free(p);}void add(char *p,int i){    //添加int l=strlen(p);tree *a=root,*b;for(int i=0;i<l;i++){int k=p[i]-'0';if(a->child[k]!=0) a=a->child[k];else{b=(tree *)malloc(sizeof(tree));for(int j=0;j<=9;j++) b->child[j]=NULL;a->child[k]=b;a=b;}}a->num=i;}void search_word(char *p){  //查找    int l=strlen(p);tree *a=root;for(int i=0;i<l;++i){        int k=sign[p[i]];if(a->child[k]==NULL) return ;else a=a->child[k];}ans[a->num]++;}void init_root(){   //初始化root    root=(tree *)malloc(sizeof(tree));    for(int i=0;i<=9;i++) root->child[i]=NULL;    memset(ans,0,sizeof(ans));    clear(root);}int main(){    init();    int T,n,m,x;    scanf("%d",&T);    while(T--){        init_root();        scanf("%d %d",&n,&m);        char s[10];        for(int i=1;i<=n;++i){            scanf("%s",&s);            add(s,i);        }        for(int i=1;i<=m;++i){            scanf("%s",s);            search_word(s);        }        for(int i=1;i<=n;++i) printf("%d\n",ans[i]);    }    return 0;}/*Sample Input13 5644484674goinnightmightgnSample Output320Source*/</span>

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