Word Puzzles（字典树）

Description

Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started using table covers with word puzzles printed on them, possibly with the intent to minimise their client‘s perception of any possible delay in bringing them their order. 

Even though word puzzles may be entertaining to solve by hand, they may become boring when they get very large. Computers do not yet get bored in solving tasks, therefore we thought you could devise a program to speedup (hopefully!) solution finding in such puzzles. 

The following figure illustrates the PizzaHut puzzle. The names of the pizzas to be found in the puzzle are: MARGARITA, ALEMA, BARBECUE, TROPICAL, SUPREMA, LOUISIANA, CHEESEHAM, EUROPA, HAVAIANA, CAMPONESA. 

Your task is to produce a program that given the word puzzle and words to be found in the puzzle, determines, for each word, the position of the first letter and its orientation in the puzzle. 

You can assume that the left upper corner of the puzzle is the origin, (0,0). Furthemore, the orientation of the word is marked clockwise starting with letter A for north (note: there are 8 possible directions in total). 

20 20 10
QWSPILAATIRAGRAMYKEI
AGTRCLQAXLPOIJLFVBUQ
TQTKAZXVMRWALEMAPKCW
LIEACNKAZXKPOTPIZCEO
FGKLSTCBTROPICALBLBC
JEWHJEEWSMLPOEKORORA
LUPQWRNJOAAGJKMUSJAE
KRQEIOLOAOQPRTVILCBZ
QOPUCAJSPPOUTMTSLPSF
LPOUYTRFGMMLKIUISXSW
WAHCPOIYTGAKLMNAHBVA
EIAKHPLBGSMCLOGNGJML
LDTIKENVCSWQAZUAOEAL
HOPLPGEJKMNUTIIORMNC
LOIUFTGSQACAXMOPBEIO
QOASDHOPEPNBUYUYOBXB
IONIAELOJHSWASMOUTRK
HPOIYTJPLNAQWDRIBITG
LPOINUYMRTEMPTMLMNBO
PAFCOPLHAVAIANALBPFS
MARGARITA
ALEMA
BARBECUE
TROPICAL
SUPREMA
LOUISIANA
CHEESEHAM
EUROPA
HAVAIANA
CAMPONESA

0 15 G
2 11 C
7 18 A
4 8 C
16 13 B
4 15 E
10 3 D
5 1 E
19 7 C
11 11 H

解题思路：

输入需要查找的单词时时，为该单词建立字典树，在单词的结尾处记录该单词的编号，然后对整个图进行暴搜，以图中的每个点作为起点，从八个方向分别查找。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
const int maxn = 26;
char map[1005][1005], word[1005][1005];
int ans[1005][3];  // 记录三个答案，分别为坐标x,y,和方向
int l, c, w;
bool visited[1005];
int dir[8][2] = {-1,0,-1,1,0,1,1,1,1,0,1,-1,0,-1,-1,-1};  //八个方向从北开始按顺势针排序
struct Trie
{
    int n;
    Trie *next[maxn];
};
Trie *root;

void init()
{
    root = (Trie *)malloc(sizeof(Trie));
    for(int i = 0; i < maxn; i++)
    {
        root -> next[i] = NULL;
        root -> n = -1;  // 建立新指针的时候，n都赋为-1，这样所有n只能为-1或编号两种可能
    }
}

void insert(char *word, int id)
{
    Trie *temp = root;
    for(int i = 0; i < strlen(word); i++)
    {
        int pos = word[i] - 'A';
        if(temp -> next[pos] == NULL)
        {
            Trie *cur = (Trie *)malloc(sizeof(Trie));
            for(int j = 0; j < maxn; j++)
            {
                cur -> next[j] = NULL;
                cur -> n = -1;
            }
            temp -> next[pos] = cur;
        }
        temp = temp -> next[pos];
    }
    temp -> n = id;  // 单词结尾处记录单词编号
}
void search(int x,int y,int d)
{
    Trie *temp = root;
    int xx = x, yy = y;
    while(xx >= 0 && xx < l && yy >=0 && yy <c)
    {
        if(temp -> next[map[xx][yy] - 'A'] == NULL)  //如果指针为空就跳出
            break;
        else
            temp = temp -> next[map[xx][yy] - 'A'];  // 否则指针下移
        if(temp -> n != -1)   // 一直找到单词的末尾处
        {
            if(!visited[temp -> n])  // 已访问过就不需要记录答案
            {
                visited[temp -> n] = 1;
                ans[temp -> n][0] = x;
                ans[temp -> n][1] = y;
                ans[temp -> n][2] = d;
            }
        }
        xx += dir[d][0];  // 朝某一方向搜索
        yy += dir[d][1];
    }
}
int main()
{
    scanf("%d%d%d", &l, &c, &w);
    for(int i = 0; i < l; i++)
        scanf("%s", map[i]);
    init();
    memset(visited, 0, sizeof(visited));
    for(int i = 0; i < w; i++)
    {
        scanf("%s", word[i]);
        insert(word[i], i);   // 建树
    }
    for(int i = 0; i < l; i++)  //对整个图进行暴搜
        for(int j = 0; j < c; j++)
            for(int k = 0; k < 8; k++)
                search(i,j,k);
    for(int i = 0; i < w; i++)
        printf("%d %d %c\n", ans[i][0], ans[i][1], ans[i][2] + 'A');
}

Word Puzzles（字典树）,布布扣,bubuko.com

Word Puzzles（字典树）

原文：http://blog.csdn.net/userluoxuan/article/details/38439545