POJ 1318 Word Amalgamation

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题目链接：http://poj.org/problem?id=1318

看了别人的代码觉得用了set数据结构真的好神奇，好简单

题目大意：从字典中寻找由一些字母组成的单词

代码：字典树，排序

思路：在字典的节点上保存下存入经过排序的

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;struct Str{char s[7];};struct nod{int c[29];Str s[100];int cnt;int fl;}p[600];int tot = 0;char temp[8];int New_node(int n){memset(p[tot].c, -1, sizeof(p[tot].c));p[tot].cnt = 0;return tot ++ ;}bool cmp(char a, char b){return a < b ;}bool cmp1(Str a, Str b){return strcmp(a.s, b.s) < 0;}void insert(char s[]){int now = 0;int len = strlen(s);strcpy(temp, s);sort(s, s+len);int i;for(i = 0; i < len; i ++){int a = s[i] - 'a';if(p[now].c[a] == -1)p[now].c[a] = New_node(i+1);now = p[now].c[a];}strcpy(p[now].s[p[now].cnt].s, temp);p[now].cnt ++;}void find(char s[]){int now = 0;int len = strlen(s);sort(s, s+len);int i;for(i = 0; i < len; i ++){int a = s[i] - 'a';if(p[now].c[a] == -1){printf("NOT A VALID WORD\n");return ;}now = p[now].c[a];}sort(p[now].s, p[now].s + p[now].cnt, cmp1);for(i = 0; i < p[now].cnt; i ++)printf("%s\n", p[now].s[i].s);}int main(){int root = New_node(0);char s[12];while(scanf("%s", s)){if(s[0] == 'X')break;insert(s);}while(scanf("%s", s)){if(s[0] == 'X')break;find(s);printf("******\n");}return 0;}

单词，然后把节点所有的单词排序，用find函数搜索到节点输出单词