[UVA][字串搜] 422 - Word-Search Wonder＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2012-05-11 07:26:21| 人1,069| 回0 | 上一篇 | 下一篇

[UVA][字串搜] 422 - Word-Search Wonder

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Word-Search Wonder

The Pyrates Restaurant was starting to fill up as Valentine McKee walked in. She scanned the crowd for her sister, brother-in-law, and nephew. Seeing her sister waving from the far end of the restaurant, she made her way back to their booth. ``Hi, Valentine,'' her sister and brother-in-law, Niki and Dennis Chapman, greeted her.

``Hi, guys,'' she replied. ``What are you doing, Wade?'' she asked her nephew. He was busy working on one of the restaurant's activity sheets with a crayon.

``I'm doing a word search game,'' Wade explained. ``I have to find all of these words in this big mess of letters. This is really hard.'' Wade looked intently at the paper in front of him.

``Can I help?'' asked Valentine, looking across the table at the activity sheet.

``Sure. These are the words we're looking for. They're the names of different kinds of Planes, Trains, and Automobiles.''

Input

The first line of input will specify the length (in characters) of the sides of the letter matrix (the matrix of letters will be square). The length, l, will be in the range . The next l lines of input will be the matrix itself, each line will contain l uppercase letters.

A list of words will follow. Each word will be on a line by itself; there will be 100 or fewer words. Each word will be 100 or fewer characters long, and will only contain uppercase letters.

The final line of input will contain a single zero character.

Output

Your program should attempt to find each word from the word list in the puzzle. A word is ``found'' if all the characters in the word can be traced in a single (unidirectional) horizontal, vertical, or diagonal line in the letter matrix. Words may not ``wrap around'' rows or columns, but horizontal and diagonal words may proceed from right to left (``backwards''). For each word that is found, your program should print the coordinates of its first and last letters in the matrix on a single line, separated by a single space. Coordinates are pairs of comma-separated integers (indexed from 1), where the first integer specifies the row number and the second integer specifies the column number.

If a word is not found, the string ``Not found'' should be output instead of a pair of coordinates.

Each word from the input can be ``found'' at most once in the puzzle.

Sample Input

5 EDEEE DISKE ESEEE ECEEE EEEEE DISC DISK DISP 0

Sample Output

1,2 4,2 2,1 2,4 Not found

#include <stdio.h>
#include <string.h>
int n, i, j;
int find(char str[], char map[][105]) {
    static int D[][2] = {{0,1},{1,0},{-1,0},{0,-1},
                        {1,1},{1,-1},{-1,1},{-1,-1}};
    int i, j, k, x, y, idx;
    for(i = 0; i < n; i++) {
        for(j = 0; j < n; j++) {
            for(k = 0; k < 8; k++) {
                x = i, y = j, idx = 0;
                while(map[x][y] == str[idx]) {
                    idx++;
                    if(str[idx] == '\0') {
                        printf("%d,%d %d,%d\n", i+1, j+1, x+1, y+1);
                        return 1;
                    }
                    x += D[k][0], y += D[k][1];
                    if(x < 0 || y < 0 || x >= n || y >= n)
                        break;
                }
            }
        }
    }
    return 0;
}
int main() {
    while(scanf("%d", &n) == 1) {
        char map[105][105], str[105];
        for(i = 0; i < n; i++) {
            scanf("%s", &map[i]);
        }
        while(scanf("%s", str) == 1) {
            if(!strcmp(str, "0"))
                break;
            if(!find(str, map))
                puts("Not found");
        }
    }
    return 0;
}