HDU 1875 Word Puzzle

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链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1857

题目：

Word Puzzle

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 427 Accepted Submission(s): 76

Problem Description

Did you heard of a little game named "Word Puzzle" ? If you didn't, what a pity !
In the game, you will be given a rectangular grid of letters, in which several words are hidden. Each word may begin anywhere in the puzzle, and may be oriented in any straight line horizontally, vertically, or diagonally. However, the words must all go down, right, or down-right. A dictionary is also given to you, indicating the words to be found in the grid.

You task is to find the locations of each word within the grid.

Input

There is only one test case.

The first line is two integers R and C separated by a whitespace. R (20 ≤ R ≤ 500) is the number of rows of the grid. C (20 ≤ C ≤ 500) is the number of columns of the grid.

The following R lines, each line will contains exactly C characters without anything else. Each character is in the range 'A' - 'Z'.

A blank line will be followed after the grid.

The following lines, each line contains a unique word in the dictionary. Each word will contain between 1 and 20 characters ( also in the range 'A' - 'Z'). The dictionary consists of at most 10000 words.

-1 means the end of dictionary.

Output

For each word, output the "ROW COL"(quotes for clarity) pair, where ROW is the 0-based row in which the first letter of the word is found, and COL is the 0-based column in which the first letter of the word is found. If the same word can be found more than once, the location in the lowest-indexed row should be returned. If there is still a tie, return the location with the lowest-indexed column. If a word cannot be found in the grid, return "-1 -1" for the word.

Sample Input

  3 5
HENRY
GAVIN
MAGICHENRY
HGM
HAG
MAVIN
-1

Sample Output

Author

XrtGavin@TJU

Source

HDU 2007 Programming Contest - Final

题目大意：

给一个字母矩阵，矩阵里可以按照横着，竖着，斜着（左上--->右下方向），组成单词。

然后询问一些单词，是否存在矩阵中。如果存在，输出它的起点位置（开始位置在越上面，越左边越好）。

分析与总结：

做这题想了足足有两天啊，我容易吗我

1. 初看，觉得水题一枚，直接枚举矩阵的起点，然后把矩阵中所有可能的单词都进行建Trie树，在Trie中要保存起点。然后很开心的提交了，结果Memory Limit Exceeded

2. 然后就一直想啊想啊想.....无线纠结中

3. 终于，我头脑上面的灯泡亮了一下，发现询问的单词只有1W个，那么我们其实就可以把询问单词建成一个Trie树，然后再枚举矩阵里面的单词，看枚举的单词是否在Trie中，在的话就把位置保存下来。

全部枚举完之后，在输出在Trie中的询问单词所保存的位置信息。

4. 之后再看看ACM官方标志图片上的那个闪亮的灯泡, 似乎有所感悟...做ACM追求的不就是灯泡亮的那一刻吗？

如果做ACM题，灯泡都不亮一下就做出来了，乐趣就少了很多。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;const int KIND = 26;
const int MAXN = 1000000;
int cnt_node;
int R,C;
char map[505][505];
char word[10005][22];struct node{bool isword;int r,c;node* next[KIND];void init(){r=c=-1;isword=false;memset(next, 0, sizeof(next));}
}Heap[MAXN];inline node* new_node(){Heap[cnt_node].init();return &Heap[cnt_node++];
}
// 把询问的单词建Trie树
void insert(node* root, char *str){for(char *p=str; *p; ++p){int ch=*p-'A';if(root->next[ch]==NULL)root->next[ch] = new_node();root=root->next[ch];} root->isword=true;
}
// 对矩阵中枚举的单词查找是否再Trie树中
void search(node* root, char *str, int row, int col){for(char *p=str; *p; ++p){int ch=*p-'A';if(root->next[ch]==NULL)return ;root=root->next[ch];if(root->isword && root->r==-1 && root->c==-1){root->r=row, root->c=col; }}if(root->isword && root->r==-1 && root->c==-1){root->r=row, root->c=col;}
}
// 输出询问的单词在矩阵中的位置
void output(node* root, char *str){for(char *p=str; *p; ++p){int ch=*p-'A';if(root->next[ch]==NULL) return;root=root->next[ch];}if(root->isword)printf("%d %d\n", root->r, root->c);
}int main(){// Trie initcnt_node=0;node* root=new_node();scanf("%d%d%*c",&R,&C);for(int i=0; i<R; ++i){gets(map[i]);}gets(word[0]); //消除空格int pos=0;while(gets(word[pos])){if(word[pos][0]=='-') break;insert(root, word[pos++]);}char str[30];for(int i=0; i<R; ++i){for(int j=0; j<C; ++j){int end_r,end_c=j;// 竖下if(i+20<R)end_r=i+20;else end_r=R;memset(str, 0, sizeof(str));for(int k=i,p=0; k<end_r; ++k)str[p++]=map[k][j];search(root, str, i, j);// 横着if(j+20<C) end_c=j+20;elseend_c=C;memset(str, 0, sizeof(str));for(int k=j,p=0; k<end_c; ++k)str[p++]=map[i][k];search(root, str, i, j);// 斜着int r=i, c=j,p=0;memset(str, 0, sizeof(str));while(r<end_r && c<end_c){str[p++]=map[r][c];++r, ++c;}search(root,str,i,j);}}for(int i=0; i<pos; ++i){output(root, word[i]);}return 0;
}

—— 生命的意义，在于赋予它意义士。

原创 http://blog.csdn.net/shuangde800 ， By D_Double (转载请标明)