HDU 1247

来源：互联网发布：奥卡姆剃刀知乎编辑：程序博客网时间：2024/05/20 12:47

思路：年前学得Trie树，当时没过，可能当时没完全理解吧，今天碰到了，就凭着理解直接敲的，调了两次终于过了，把每个单词结尾标记为1,查询的时候看某个单词的子串是否存在，这题WA点不少，必须搞清判断条件。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespace std;class Node{    public:        int cnt;        Node *child[26];        Node(){            cnt = 0;            for(int i = 0;i < 26;i ++) child[i] = NULL;        }};Node *root = new Node();char str[50005][111];void Update(char *str){    Node *tmp = root;    while(*str != '\0'){        if(tmp->child[*str-'a'] == NULL){            Node *p = new Node();            tmp->child[*str-'a'] = p;        }        tmp = tmp->child[*str-'a'];        str++;    }    tmp->cnt = 1;}bool Judge(char *str){    Node *tmp = root;    while(*str != '\0'){        if(tmp->child[*str-'a'] == NULL) return false;        tmp = tmp->child[*str-'a'];        str++;    }    return tmp->cnt == 1;}int main(){    int n = 0;//    freopen("in.cpp","r",stdin);    memset(str,0,sizeof(str));    while(~scanf("%s",str[n])) Update(str[n]),n++;    for(int i = 0;i < n;i ++){        int len = strlen(str[i]);        char tmp1[111],tmp2[111];        memset(tmp1,0,sizeof(tmp1));        memset(tmp2,0,sizeof(tmp2));        for(int j = 1;j < len;j ++){            int p = 0;            for(int k = 0;k < j;k ++) tmp1[p++] = str[i][k];            p = 0;            for(int k = j;k < len;k ++) tmp2[p++] = str[i][k];            if(Judge(tmp1) && Judge(tmp2)) {                printf("%s\n",str[i]);                break;            }            memset(tmp1,0,sizeof(tmp1));            memset(tmp2,0,sizeof(tmp2));        }    }    return 0;}

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