c - 如何在二维字符数组中搜索相同的字段？(在线游戏中的五个)在 C 中

Question

我是 C 和编程的新手，我的家庭作业有问题。 邹氏都知道游戏叫“五排”。我的工作是创建一个程序，在每次移动后评估玩家的位置。它必须说明哪条线最长(垂直、水平和对角线方向)，并且还打印字段中有多少条这样的线。 所以输入是位置，输出应该是这样的:玩家“X”有最长的线 3(3 次)。 我有一个这样的字段:

#define MAX_LINE 26
#define MAX_COLUMN 26

char playField[MAX_LINE][MAX_COLUMN]

经过几步之后，我遇到了这样的情况:

playField[1][2]=='X'
playField[1][3]=='O'
playField[2][2]=='X'
playField[1][4]=='O'
playField[9][5]=='X'

那么如何搜索最长的行呢？这只是一个例子，字段可以有更多的元素。 我的想法是用一些周期来做，但似乎不可能为我写出来。 感谢帮助

Answer 1

int maxO = 0; //max length of O's
int maxX = 0; //max length of X's


char last;     //the last player we checked (or nothing)
int  lastCount;//the last count for the "last"

检查线

//lines along lines
/*
    _ _ _ _ _ _ _
    _ _ X X X _ _
    _ _ _ _ _ _ _

    for each line 
        for each column
            check for longest X/0

*/
for(int i = 0; i < MAX_LINE; ++i) {
    last = 0;
    lastCount = 0;
    for(int j = 0; j < MAX_COLUMN; ++j) {
        assert(i >= 0 && i < MAX_LINE && j >= 0 && j < MAX_COLUMN);
        //check if same player
        if (last == playField[i][j]) {
            //same player
            lastCount += 1;
        } else {
            //not same player, set counted length for the right player
            if (last == 'O' && maxO < lastCount) {
                //it's player O
                maxO = lastCount;
            } else if (last  == 'X' && maxX < lastCount) {
                //it's player X
                maxX = lastCount;
            }
            //update for next player (or nothing)
            lastCount = 1;
            last = playField[i][j];
        }
    }
    if (last == 'O' && maxO < lastCount) {
        //it's player O
        maxO = lastCount;
    } else if (last  == 'X' && maxX < lastCount) {
        //it's player X
        maxX = lastCount;
    }
}

检查列

//lines along columns
/*
    _ _ _ X _ _ _
    _ _ _ X _ _ _
    _ _ _ X _ _ _

    for each column
        for each line
            check for longest X/0

*/
for(int j = 0; j < MAX_COLUMN; ++j) {
    last = 0;
    lastCount = 0;
    for(int i = 0; i < MAX_LINE; ++i) {
        assert(i >= 0 && i < MAX_LINE && j >= 0 && j < MAX_COLUMN);
        if (last == playField[i][j]) {
            lastCount += 1;
        } else {
            if (last == 'O' && maxO < lastCount) {
                maxO = lastCount;
            } else if (last  == 'X' && maxX < lastCount) {
                maxX = lastCount;
            }
            lastCount = 1;
            last = playField[i][j];
        }
    }
    if (last == 'O' && maxO < lastCount) {
        //it's player O
        maxO = lastCount;
    } else if (last  == 'X' && maxX < lastCount) {
        //it's player X
        maxX = lastCount;
    }
}

检查对角线

//lines along diagonals
/*
    _ _ _ X _ _ _
    _ _ X _ _ _ _
    _ X _ _ _ _ _

    for each line + column
        for each column
            shift coordinates
                0 1 2 3 4 5 6
                _ _ _ X _ _ _
                _ _ X _ _ _ _
                _ X _ _ _ _ _
                to
                0 1 2 3 4 5 6 7 8
                _ _ _ X _ _ _
                  _ _ X _ _ _ _
                    _ X _ _ _ _ _
              check if shifted coordiantes are in range
                  check for longest X/0

*/
for(int i = 0; i < MAX_LINE + MAX_COLUMN - 1; ++i) {
    last = 0;
    lastCount = 0;
    for(int j = 0; j < MAX_COLUMN; ++j) {
        int iShift = i - j;
        if (iShift >= 0 && iShift < MAX_LINE) {
            assert(iShift >= 0 && iShift < MAX_LINE && j >= 0 && j < MAX_COLUMN);
            if (last == playField[iShift][j]) {
                lastCount += 1;
            } else {
                if (last == 'O' && maxO < lastCount) {
                    maxO = lastCount;

                } else if (last  == 'X' && maxX < lastCount) {
                    maxX = lastCount;
                }
                lastCount = 1;
                last = playField[iShift][j];
            }
        }
    }
    if (last == 'O' && maxO < lastCount) {
        //it's player O
        maxO = lastCount;
    } else if (last  == 'X' && maxX < lastCount) {
        //it's player X
        maxX = lastCount;
    }
}

检查对角线 2

//lines along diagonals 2
/*
    _ _ _ X _ _ _
    _ _ _ _ X _ _
    _ _ _ _ _ X _

    for each line + column
        for each column
            shift coordinates
                0 1 2 3 4 5 6
                _ _ _ X _ _ _
                _ _ _ _ X _ _
                _ _ _ _ _ X _

                to
               -2-1 0 1 2 3 4 5 6
                    _ _ _ X _ _ _
                  _ _ _ _ X _ _
                _ _ _ _ _ X _

              check if shifted coordiantes are in range
                  check for longest X/0

*/
for(int i = -MAX_COLUMN + 1; i < MAX_LINE; ++i) {
    last = 0;
    lastCount = 0;
    for(int j = 0; j < MAX_COLUMN; ++j) {
        int iShift = i + j;            
        if (iShift >= 0 && iShift < MAX_LINE) {
            assert(iShift >= 0 && iShift < MAX_LINE && j >= 0 && j < MAX_COLUMN);
            if (last == playField[iShift][j]) {
                lastCount += 1;
            } else {
                if (last == 'O' && maxO < lastCount) {
                    maxO = lastCount;
                } else if (last  == 'X' && maxX < lastCount) {
                    maxX = lastCount;
                }
                lastCount = 1;
                last = playField[iShift][j];
            }
        }
    }
    if (last == 'O' && maxO < lastCount) {
        //it's player O
        maxO = lastCount;
    } else if (last  == 'X' && maxX < lastCount) {
        //it's player X
        maxX = lastCount;
    }
}

打印结果

//print result
printf("Player X longest line %d\n", maxX);
printf("Player O longest line %d\n", maxO);