在给定二维字符矩阵中查找给定字符串

具体题目描述见:http://www.careercup.com/question?id=13126665

对于这个题目我想到的方法是暴力搜索，因为目标串已经给定，如果我们将矩阵中每个属于目标串的字符看做图的结点，如果目标串中出现形如“AB”的子串且A和B在字符矩阵中相邻，则在建立一条由A到B的有向边。我们只需要搜索一条长度和目标串长度相同的路径即可。代码如下：

#include <stdio.h>
#define M 100
#define N 50

char str[M + 1][M + 1];
char pat[N + 1];
int lp;
char flag[M + 1][M + 1];
int n, m;

int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};

int do_dfs(int x, int y, int idx) {
    int i, tx, ty;
    if (idx == lp - 1) {
        return 1;
    }
    else {
        for (i = 0; i < 8; i++) {
            tx = (x + dx[i] + n) % n;
            ty = (y + dy[i] + m) % m;
            if ((!flag[tx][ty]) && (str[tx][ty] == pat[idx + 1])) {
                flag[tx][ty] = 1;
                if (do_dfs(tx, ty, idx + 1)) {
                    return 1;
                }
                else {
                    flag[tx][ty] = 0;
                }
            }
        }
    }
    return 0;
}

int is_matching()
{
    int x, y, i, j;
    for (x = 0; x < n; x++) {
        for (y = 0; y < m; y++) {
            for (i = 0; i < n; i++) {
                for (j = 0; j < m; j++) {
                    flag[i][j] = 0;
                }
            }

            if (str[x][y] == pat[0]) {
                flag[x][y] = 1;
                if (do_dfs(x, y, 0)) {
                    return 1;
                }
            }
        }
    }
    return 0;
}

int main()
{
    int i;
    n = 5, m = 5;
    str[0][0] = 'a', str[0][1] = 'c', str[0][2] = 'p', str[0][3] = 'r', str[0][4] = 'c';
    str[1][0] = 'x', str[1][1] = 's', str[1][2] = 'o', str[1][3] = 'p', str[1][4] = 'c';
    str[2][0] = 'v', str[2][1] = 'o', str[2][2] = 'v', str[2][3] = 'n', str[2][4] = 'i';
    str[3][0] = 'w', str[3][1] = 'g', str[3][2] = 'f', str[3][3] = 'm', str[3][4] = 'n';
    str[4][0] = 'q', str[4][1] = 'a', str[4][2] = 't', str[4][3] = 'i', str[4][4] = 't';
    gets(pat);
    lp = strlen(pat);
    printf("%d\n", is_matching());
    return 0;