zoj 1109 Language of FatMouse

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点击打开zoj 1190

题意：给定一序列的映射关系，然后再给定一些单词，要求如果该单词有映射的单词就输出映射的单词，否则输出"eh"

思路：把给定的映射关系的中的单词建立成字典树，然后输入单词的时候查找即可。但是这一题不能够利用静态分配的思想，应该要利用动态的分配

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN = 100010;const int N = 11;const int M = 26;struct Node{    int index;    Node* child[M];    Node(){        index = -1;        for(int i = 0 ; i < M ; i++)           child[i] = NULL;    }};Node* root;int pos;char word[MAXN][N];void insert(int index , char* str){    Node* s = root;    int len = strlen(str);    for(int i = 0 ; i < len ; i++){        int num = str[i]-'a';        if(s->child[num] == NULL)            s->child[num] = new Node();        s = s->child[num];    }    s->index = index;}int search(char *str){    Node* s = root;    int len = strlen(str);    for(int i = 0 ; i < len ; i++){        int num = str[i]-'a';        if(s->child[num] == NULL)            return -1;        s = s->child[num];    }    return s->index;}int main(){    pos = 0;    root = new Node();    char str[N*3];    char tmp[N];    int cnt = 0;        while(gets(str) && str[0] != '\0'){        sscanf(str , "%s %s" , word[cnt] , tmp);        insert(cnt++ , tmp);    }    while(scanf("%s" , tmp) != EOF){        int index = search(tmp);        if(index == -1)            printf("eh\n");        else            printf("%s\n" , word[index]);    }    return 0;}