HDU 1247 Hat's words（Trie）

ACM

题目地址： 
HDU 1247 Hat's words

题意： 
给些单词，问每个单词是否能用另外两个单词拼出。

分析： 
直接保存到trie里面，然后暴力切割查询即可。

代码：

/**  Author:      illuz <iilluzen[at]gmail.com>*  File:        1247.cpp*  Create Date: 2014-09-24 11:04:11*  Descripton:   */#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)typedef long long ll;const int N = 50010;const int MAXNODE = 1000010;const int MAXSON = 26;// array indexstruct ATrie {int ch[MAXNODE][MAXSON];int val[MAXNODE];int sz;// num of nodesATrie() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); }void init() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); }inline int idx(char c) { return c ? c - 'a' : 0; }void insert(char *s, int v = 1) {int u = 0, len = strlen(s);repf (i, 0, len - 1) {int c = idx(s[i]);if (!ch[u][c]) {memset(ch[sz], 0, sizeof(ch[sz]));val[sz] = 0;ch[u][c] = sz++;}u = ch[u][c];}val[u] = v;}// if s in trie return the value, else return 0int find(char *s) {int u = 0, len = strlen(s);repf (i, 0, len - 1) {int c = idx(s[i]);if (ch[u][c])u = ch[u][c];elsereturn 0;}return val[u];}} trie;char wd[N][110], a[110], b[110];int n;int main() {// ios_base::sync_with_stdio(0);while (gets(wd[n])) {trie.insert(wd[n++]);}repf (i, 0, n - 1) {int len = strlen(wd[i]);repf (j, 1, len - 1) {strncpy(a, wd[i], j);a[j] = 0;strcpy(b, wd[i] + j);// cout << a << ' ' << b << endl;if (trie.find(a) && trie.find(b)) {printf("%s\n", wd[i]);break;}}}return 0;}