The Tamworth Two

题意：有一个10×10的网格，网格的每一小格可以是障碍（*），空地（.），农夫（F）或者牛（C）。农夫和牛都有四个朝向，东南西北，初始情况下都朝北。二者都按照下面的规则来运动：如果向前走一步不是障碍或者走出网格，则向前走。否则顺时针改变朝向90度。每一分钟农夫和牛同时运动一步，求什么时候农夫和牛能够相遇在同一格。

解题思路：

1. 读入网格的状态，用数组maze[i][j]来保存，初始化为0，读到障碍的时候设为1。在读取过程中保存农夫和牛的初始状态。
2. 农夫和牛的运动一定是有规律的，也就是在若干步（可以是0）以后陷入一个循环。
3. 模拟农夫的运动，用farmer_state[i][j][k]来保存状态，全部初始化为-1。farmer_state[i][j][k] = x，代表在x分钟的时候，农夫走到坐标（i, j）处，并且朝向为k（北东南西分别为0,1,2,3）。
4. 在某一时刻会发现farmer_state[i][j][k] != -1，那么就说明回到了前面的状态，此时农夫开始陷入循环，而且循环的长度就是此时的时间减去前面状态记录的时间。这样在模拟过程中可以记录农夫在何时陷入循环（farmer_repeat_from）和这个循环的长度（farmer_repeat_length）。
5. 得到上述的farmer_state之后对其进行遍历，得到一个新的数组farmer_route[i][j]，其中j取0和1。farmer_route[i][0]和farmer_route[i][1]分别代表农夫在经过i步之后所在的网格坐标(x, y)中的x和y。
6. 对牛重复上述3~5操作即可得到相应的cows_state，cows_repeat_from，cows_repeat_length，cows_route。
7. 求farmer_repeat_length和cows_repeat_length的最小公倍数，将其加上farmer_repeat_from和cows_repeat_from中较大的值，得到end_point。
8. 对[0, end_point)这些时刻进行遍历，直到找到一个农夫和牛都处于同一网格的时刻（从farmer_route和cows_route中查找，相应的index用XXX_repeat_from和XXX_repeat_length计算得到）。end_point之后的时刻不需要进行处理，因为后面又会陷入另外一个循环（农夫与牛二者相对状态的循环）。

代码：

/*ID: zc.rene1LANG: CPROG: ttwo*/#include<stdio.h>#include<stdlib.h>#include<string.h>enum directions{NORTH=0, EAST, SOUTH, WEST};struct unit{    int i, j;    int direction;};struct unit farmer, cows;int maze[12][12];int farmer_state[12][12][4];int cows_state[12][12][4];int GetCommonFactor(int a, int b){    while ((a != 0)&&(b != 0))    {if (a > b){    a %= b;}else{    b %= a;}    }    return (a==0)? b:a;}void Move(struct unit *in){    int new_i = (*in).i;    int new_j = (*in).j;    switch((*in).direction)    {case 0:    new_i -= 1;    break;case 1:    new_j += 1;    break;case 2:    new_i += 1;    break;case 3:    new_j -= 1;    break;default:    break;    }    if (maze[new_i][new_j] == 0)    {if ((*in).direction == 3){    (*in).direction = 0;}else{    (*in).direction += 1;}    }    else    {(*in).i = new_i;(*in).j = new_j;    }}int main(void){    FILE *fin, *fout;    int i, j, k, count, temp;    char *str = NULL;    char lt;    int **farmer_route, **cows_route;    fin = fopen("ttwo.in", "r");    fout = fopen("ttwo.out", "w");    str = (char*)malloc(11*sizeof(char));    memset(maze, 0, 12*12*sizeof(int));    memset(farmer_state, -1, 12*12*4*sizeof(int));    memset(cows_state, -1, 12*12*4*sizeof(int));    for (i=1; i<11; i++)    {fscanf(fin, "%s", str);for (j=1; j<11; j++){    lt = str[j-1];    if (lt != '*')    {maze[i][j] = 1;    }    if (lt == 'F')    {farmer.i = i;farmer.j = j;farmer.direction = NORTH;    }    if (lt == 'C')    {cows.i = i;cows.j = j;cows.direction = NORTH;    }}    }    count = 0;    while (farmer_state[farmer.i][farmer.j][farmer.direction] == -1)    {farmer_state[farmer.i][farmer.j][farmer.direction] = count;count++;Move(&farmer);    }    int farmer_repeat_from = farmer_state[farmer.i][farmer.j][farmer.direction];    int farmer_repeat_length = count - farmer_repeat_from;    farmer_route = (int **)malloc(count*sizeof(int*));    for (i=0; i<count; i++)    {farmer_route[i] = (int *)malloc(2*sizeof(int));memset(farmer_route[i], 0, 2*sizeof(int));    }    for (i=1; i<11; i++)    {for (j=1; j<11; j++){    for (k=0; k<4; k++)    {temp = farmer_state[i][j][k];if (temp != -1){    farmer_route[temp][0] = i;    farmer_route[temp][1] = j;}    }}    }    count = 0;    while (cows_state[cows.i][cows.j][cows.direction] == -1)    {cows_state[cows.i][cows.j][cows.direction] = count;count++;Move(&cows);    }    int cows_repeat_from = cows_state[cows.i][cows.j][cows.direction];    int cows_repeat_length = count - cows_repeat_from;    cows_route = (int **)malloc(count*sizeof(int*));    for (i=0; i<count; i++)    {cows_route[i] = (int *)malloc(2*sizeof(int));memset(cows_route[i], 0, 2*sizeof(int));    }    for (i=1; i<11; i++)    {for (j=1; j<11; j++){    for (k=0; k<3; k++)    {temp = cows_state[i][j][k];if (temp != -1){    cows_route[temp][0] = i;    cows_route[temp][1] = j;}    }}    }    int greatest_common_factor = GetCommonFactor(farmer_repeat_length, cows_repeat_length);    int greatest_common_multiple = farmer_repeat_length * cows_repeat_length / greatest_common_factor;    int end_point = ((farmer_repeat_from > cows_repeat_from)?farmer_repeat_from:cows_repeat_from) + greatest_common_multiple;    int farmer_index, cows_index;    for (i=0; i<end_point; i++)    {if (i < farmer_repeat_from){    farmer_index = i;}else{    farmer_index = ((i - farmer_repeat_from) % farmer_repeat_length) + farmer_repeat_from;}if (i < cows_repeat_from){    cows_index = i;}else{    cows_index = ((i - cows_repeat_from) % cows_repeat_length) + cows_repeat_from;}if ((farmer_route[farmer_index][0] == cows_route[cows_index][0])&&(farmer_route[farmer_index][1] == cows_route[cows_index][1])){    break;}    }    if (i == end_point)    {fprintf(fout, "0\n");    }    else    {fprintf(fout, "%d\n", i);    }    return 0;}