hdu 1247 （字典树）

来源：互联网发布：linux挂载u盘ntfs 编辑：程序博客网时间：2024/06/05 19:53

点击打开链接

分析：

对每一个字符串进行分解，看它分解的两部分能否在字典树中find，用hash标记。。。

WA了好几遍。。。

下面是大神的ice_Crazy的代码。。

#include"stdio.h"#include"stdlib.h"#include"string.h"struct dictree{    struct dictree *child[26];    int flag;};struct dictree *root;char str[50011][50];void insert(char *source){    int i,j;    int len;    struct dictree *current,*newnode;    len=strlen(source);    current=root;    for(i=0;i<len;i++)    {        if(current->child[source[i]-'a']!=0)            current=current->child[source[i]-'a'];        else        {            newnode=(struct dictree *)malloc(sizeof(struct dictree));            for(j=0;j<26;j++)    newnode->child[j]=0;            newnode->flag=0;            current->child[source[i]-'a']=newnode;            current=newnode;        }    }    current->flag=1;}int find(char *source){    int i;    int len;    struct dictree *current;    len=strlen(source);    current=root;    for(i=0;i<len;i++)    {        if(current->child[source[i]-'a']!=0)            current=current->child[source[i]-'a'];        else    return 0;    }    return current->flag;}int main(){    char t1[50],t2[50];    int t;    int k;    int i,l,j,j2;    int len;    int hash[50011];    root=(struct dictree *)malloc(sizeof(struct dictree));    for(j=0;j<26;j++)    root->child[j]=0;    root->flag=0;    k=0;    while(scanf("%s",str[k])!=-1)    {        insert(str[k]);        k++;    }    for(i=0;i<k;i++)    {        len=strlen(str[i]);        hash[i]=0;        if(len==1)    continue;                t=len-1;        for(l=1;l<=t;l++)        {            for(j=0;j<l;j++)        t1[j]=str[i][j];            t1[j]=0;            for(j2=0;j<len;j++,j2++)t2[j2]=str[i][j];            t2[j2]=0;            if(find(t1)&&find(t2))    {hash[i]=1;break;}        }    }    for(i=0;i<k;i++)    if(hash[i])    printf("%s\n",str[i]);    return 0;}

我的代码：（还没A呢）

#include"stdio.h"#include"string.h"#include"stdlib.h"struct tree{struct tree *child[26];int num;};struct tree *root;void insert(char *p){int i,j;int len;struct tree *cur,*nee;cur=root;len=strlen(p);for(i=0;i<len;i++){if(cur->child[p[i]-'a']!=0){cur=cur->child[p[i]-'a'];//cur->num++;}else{nee=(struct tree*)malloc(sizeof(struct tree));for(j=0;j<26;j++)nee->child[j]=0;nee->num=0;cur->child[p[i]-'a']=nee;cur=nee;}}cur->num=1;}int find(char *p){int i;int len;struct tree *cur;cur=root;len=strlen(p);for(i=0;i<len;i++){if(cur->child[p[i]-'a']!=0)cur=cur->child[p[i]-'a'];else return 0;}return cur->num;}int hash[50011];char str[50011][51];int main(){int i,j,l,k;int t1,t2,len;char s1[55],s2[55];root=(struct tree*)malloc(sizeof(struct tree));root->num=0;for(i=0;i<26;i++)root->child[i]=0;k=0;while(gets(str[k]))insert(str[k++]);for(i=0;i<k;i++){hash[i]=0;len=strlen(str[i]);if(len==1)continue;for(j=1;j<len;j++){t1=t2=0;for(l=0;l<j;l++)s1[t1++]=str[i][l];s1[t1]=0;for(;l<len;l++)s2[t2++]=str[i][l];s2[t2]=0;if(find(s1)&&find(s2)){hash[i]=1;break;}}}for(i=0;i<k;i++){if(hash[i])printf("%s\n",str[i]);}return 0;}