Judgment Problem of Gobang in Machine Test

Two mainstream ideas:

1. Scan the board directly and traverse one by one in four directions, which is not only troublesome but also silly;

2. Each time a child is dropped, the four directions of the child are directly judged, and the left and right directions can be summed up in one direction;

 

#include<iostream>
#include<vector>
#include<string>
using namespace std;

const  int maxn = 13 ;
int ma [maxn] [maxn];

void printma() {
    for (int i = 0; i < maxn; i++) {
        for (int j = 0; j < maxn; j++) {
            if (ma[i][j] == 1) {
                cout << "o";
            }
            else if (ma[i][j] == -1) {
                cout << "x";
            }
            else {
                cout << "#";
            }
            cout << " ";
        }
        cout << endl;
    }
}

int henc(int x, int y) {
    int cnt = 0;
    for (int i = x; i < maxn&&ma[x][y]==ma[i][y]; i++) {
        cnt++;
    }
    for (int i = x - 1; i >= 0 && ma[x][y] == ma[i][y]; i--)
        cnt++;
    return cnt;
}

int shuc(int x, int y) {
    int cnt = 0;
    for (int i = y; i < maxn&&ma[x][y] == ma[x][i]; i++) {
        cnt++;
    }
    for (int i = y - 1; i >= 0 && ma[x][y] == ma[x][i]; i--)
        cnt++;
    return cnt;
}

int xzc(int x, int y) {
    int cnt = 0;
    for (int i = x, j = y; i < maxn&&y < maxn&&ma[x][y] == ma[i][j]; i++, j++) {
        cnt++;
    }
    for (int i = x-1, j = y-1; i >=0&&y >= 0&&ma[x][y] == ma[i][j]; i--, j--) {
        cnt++;
    }
    return cnt;
}

int xyc(int x, int y) {
    int cnt = 0;
    for (int i = x, j = y; i >=0&&y < maxn&&ma[x][y] == ma[i][j]; i--, j++) {
        cnt++;
    }
    for (int i = x, j = y; i < maxn && y >= 0 &&ma[x][y] == ma[i][j]; i++, j--) {
        cnt++;
    }
    return cnt;
}

bool charge(int x,int y) {
    if (henc(x, y) == 5 || shuc(x, y) == 5 || xzc(x, y) == 5 || xyc(x, y) == 5)
        return true;
    else
        return false;
}

int main() {
    int x, y;
    int f = 1;
    while (1) {
        cin >> x >> y;
        has [x] [y] = f;
        printma ();
        if (charge(x, y)) {
            if (f == 1) {
                cout << " Black o victory " << endl;
            }
            else {
                cout << " White x victory " << endl;
            }
            break;
        }
        f = -f;
    }
    return 1;
}