UVA - 10010 Where's Waldorf?

最新推荐文章于 2017-09-06 15:03:27 发布

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最新推荐文章于 2017-09-06 15:03:27 发布

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本文链接：https://blog.youkuaiyun.com/cwb15505905513/article/details/51863403

本文介绍了解决UVA-10010 Where's Waldorf?问题的一种算法实现，该算法能在给定的矩阵中从八个方向搜索目标字符串，并返回其起始位置。

UVA - 10010 Where’s Waldorf?

题目大意：给一组英文，和给一串英文，判断这串英文从一组英文中的哪个字母开始的从上下左右左上左下右上右下八个方向中的任意一个方向找到其匹配的英文

解题思路：先从那组英文中先找到一串英文的首字母然后从八个方向找就可以了

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int main () {
    bool f;
    char a[100][100], c[100][100];
    int N;
    int t = 1;
    int n, m, k;
    int x, y;
    scanf("%d", &N);
    while ( N--) {
        x = 0;
        n = 0;
        m = 0;
        k = 0;
        y = 0;
        getchar();
        scanf("%d%d", &n, &m);
        getchar();
        if (t != 1) cout<<endl;
        t++;
        for (int i = 0; i < n; i++ ) {
            for (int j = 0; j < m; j++) {
                scanf("%c", &a[i][j]); 
                if ( a[i][j] >= 'a' )
                    a[i][j] = a[i][j] - 32;
            }
            getchar();
        } 
        scanf("%d", &k);
        getchar();
        for (int i = 0; i < k; i++) {
            scanf ("%s", c[i]);
            for (int j = 0; j < strlen(c[i]); j++)
                if ( c[i][j] >= 'a')
                    c[i][j] = c[i][j] - 32;
            getchar();
        }
//      printf("%d %d %d %d\n",n ,m ,k ,N);
        for (int i = 0; i < k ; i++) {
            f = 0;
            for (int i2 = 0; i2 < n; i2++) {
                for (int j2 = 0; j2 < m; j2++) {
                    if (c[i][0] == a[i2][j2]) {
                        f = 1;
                        for (int j = 1; j < strlen(c[i]); j++) {
                            if (c[i][j] != a[i2][j2+j] || j2+j >= m) {
                                f = 0;
                                break;
                            }
                            else f = 1;
                        }
                        if (!f) 
                            for (int j = 1; j < strlen(c[i]); j++) {
                                if (c[i][j] != a[i2][j2-j] || j2 - j < 0) { 
                                    f = 0;
                                    break;
                                }
                                else f = 1;
                            }
                        if (!f)
                            for (int j = 1; j < strlen(c[i]); j++) {
                                if (c[i][j] != a[i2+j][j2] || i2 + j >= n) {    
                                    f = 0;
                                    break;
                                }
                                else f = 1;
                            }
                        if (!f)
                            for (int j = 1; j < strlen(c[i]); j++) {
                                if (c[i][j] != a[i2-j][j2] || i2 - j < 0) { 
                                    f = 0;
                                    break;
                                }
                                else f = 1;
                            }
                        if (!f)
                            for (int j = 1; j < strlen(c[i]); j++) {
                                if (c[i][j] != a[i2-j][j2-j] || j2 - j < 0 || i2 - j < 0) { 
                                    f = 0;
                                    break;
                                }
                                else f = 1;
                            }
                        if (!f)
                            for (int j = 1; j < strlen(c[i]); j++) {
                                if (c[i][j] != a[i2+j][j2+j] || j2 + j >= m || i2 + j >= n ) {  
                                    f = 0;
                                    break;
                                }
                                else f = 1;
                            }
                        if (!f)
                            for (int j = 1; j < strlen(c[i]); j++) {
                                if (c[i][j] != a[i2+j][j2-j] || j2 - j < 0 || i2 + j >= n) {    
                                    f = 0;
                                    break;
                                }
                                else f = 1;
                            }
                        if (!f)
                            for (int j = 1; j < strlen(c[i]); j++) {
                                if (c[i][j] != a[i2-j][j2+j] || j2 + j >= m || i2 - j < 0) {    
                                    f = 0;
                                    break;
                                }
                                else f = 1;
                        }
                    }
                    if (f) { y = j2 + 1; break;}
                }
                if (f) { x = i2 + 1; break;}
            }
            if (f)
                printf("%d %d\n", x, y);
        }
    }               
    return 0;
}