题目链接：

HDU: http://acm.hdu.edu.cn/showproblem.php?pid=1986

ZOJ: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1988

HDU 1987 & ZOJ 2990 和这题刚好相反，也是比较容易模拟；

Chip and Dale have devised an encryption method to hide their (written) text messages. They first agree secretly on two numbers that will be used as the number of rows (R) and columns (C) in a matrix. The sender encodes an intermediate format using the following rules:

1.

The text is formed with uppercase letters [A-Z] and <space>.

2.

Each text character will be represented by decimal values as follows:

<space> = 0, A = 1, B = 2, C = 3, ..., Y = 25, Z = 26 The sender enters the 5 digit binary representation of the characters?? values in a spiral pattern along the matrix as shown below. The matrix is padded out with zeroes (0) to fill the matrix completely. For example, if the text to encode is: "ACM" and R=4 and C=4, the matrix would be filled in as follows:

A = 00001, C = 00011, M = 01101(one extra 0) The bits in the matrix are then concatenated together in row major order and sent to the receiver. The example above would be encoded as: 0000110100101100

Input

The first line of input contains a single integer N, (1 <= N <= 1000) which is the number of datasets that follow.

Each dataset consists of a single line of input containing R (1 <= R <= 21), a space, C (1 <= C <= 21), a space, and a text string consisting of uppercase letters [A-Z] and <space>. The length of the text string is guaranteed to be <=(R*C)/5.

Output

For each dataset, you should generate one line of output with the following values: The dataset number as a decimal integer (start counting at one), a space, and a string of binary digits (R*C) long describing the encoded text. The binary string represents the values used to fill in the matrix in row-major order. You may have to fill out the matrix with zeroes (0) to complete the matrix.

Sample Input

4
4 4 ACM
5 2 HI
2 6 HI
5 5 HI HO

Sample Output

1 0000110100101100
2 0110000010
3 010000001001
4 0100001000011010110000010

学习了http://blog.csdn.net/tdreamge/article/details/7324989感觉很清晰！

代码如下：

#include <cstdio>
#include <cstring>
const int maxn = 517;
int main()
{
    char m[47][6]= {"00000","00001","00010","00011","00100","00101","00110",
                    "00111","01000","01001","01010","01011","01100","01101",
                    "01110","01111","10000","10001","10010","10011","10100",
                    "10101","10110","10111","11000","11001","11010"
                   };
    int t;
    int r, c;
    char s1[maxn];
    int a[maxn*5], b[27][27];
    int cas = 0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&r,&c);
        getchar();
        gets(s1);
        int len = strlen(s1);
        int k = 0;
        for(int i = 0; i < len; i++)
        {
            //printf("%c",s1[i]);
            for(int j = 0; j < 5; j++)
            {
                if(s1[i] == ' ')
                    a[k++] = 0;
                else
                    a[k++] = m[s1[i]-'A'+1][j] - '0';
            }
        }
        for(int i = len*5; i < r*c; i++)//不够 补零
        {
            a[k++] = 0;
        }
//        k = 0;
//        for(int i = 0; i < len; i++)
//        {
//            for(int j = 0; j < 5; j++)
//            {
//                printf("%d",a[k++]);
//            }
//            printf("\n");
//        }
        k = 1;
        int x = 0, y = 0;
        int mark[27][27];
        memset(mark,0,sizeof(mark));
        b[0][0] = a[0];
        mark[0][0] = 1;
        while(k < r*c)
        {
            while(x+1 < c && !mark[y][x+1])//向右
            {
                x++;
                b[y][x] = a[k++];
                mark[y][x] = 1;
            }
            while(y+1 < r && !mark[y+1][x])//向下
            {
                y++;
                b[y][x] = a[k++];
                mark[y][x] = 1;
            }
            while(x-1 >= 0 && !mark[y][x-1])//向左
            {
                x--;
                b[y][x] = a[k++];
                mark[y][x] = 1;
            }
            while(y-1 >= 0 && !mark[y-1][x])//向上
            {
                y--;
                b[y][x] = a[k++];
                mark[y][x] = 1;
            }
        }
        printf("%d ",++cas);
        //printf("\n");
        for(int i = 0; i < r; i++)
        {
            for(int j = 0; j < c; j++)
            {
                printf("%d",b[i][j]);
            }
            //printf("\n");
        }
        printf("\n");
    }
    return 0;
}

HDU 1986 & ZOJ 2989 Encoding（模拟）