UVA - 1589 Xiangqi

最新推荐文章于 2024-07-28 10:44:26 发布

LOUTINI

最新推荐文章于 2024-07-28 10:44:26 发布

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分类专栏： UVA

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本文提供了一种解决UVA-1589象棋问题的算法实现，通过C++代码详细展示了如何判断红方的“帅”是否被将军，并检查是否有合法的走法来解除将军状态。代码中包含了对不同棋子走法规则的实现。

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UVA - 1589 Xiangqi

#include"iostream"
#include"algorithm"
using namespace std;
char map[10][11];
struct red {
    char name;
    int x, y;
}r[8];
int n;
int num(int a,int b,char c,int dim)//计算两个同行或同列棋子中间的棋子个数，c=='x'计算同行
{
    int cnt = 0;
    int begin = min(a, b);
    int end = max(a, b);
    for (int j = begin + 1; j < end; j++)
    {
        if (c == 'x'&&map[j][dim] != 0)cnt++;
        if (c == 'y'&&map[dim][j] != 0)cnt++;
    }
    return cnt;
}
bool check(int x,int y)
{
    if (x < 4 || x>6 || y < 1 || y>3)return false;
    for (int i = 1; i <= n; i++)
    {
        if (x == r[i].x&&y == r[i].y)continue;
        if (r[i].name == 'R' || r[i].name == 'G')//车和帅
        {
            if (y == r[i].y&&num(r[i].x, x, 'x', y) == 0)return false;
            if (x == r[i].x&&num(r[i].y, y, 'y', x) == 0)return false;
        }
        if (r[i].name == 'C')//炮
        {
            if (y == r[i].y&&num(r[i].x, x, 'x', y) == 1)return false;
            if (x == r[i].x&&num(r[i].y, y, 'y', x) == 1)return false;
        }
        if (r[i].name == 'H')//马
        {
            if (map[r[i].x + 1][r[i].y] == 0)//蹩脚没蹩脚
                if (r[i].x + 2 == x && (r[i].y + 1 == y || r[i].y - 1 == y))return false;
            if (map[r[i].x][r[i].y - 1] == 0)
                if ((r[i].x + 1 == x || r[i].x - 1 == x) && r[i].y - 2 == y)return false;
            if (map[r[i].x - 1][r[i].y] == 0)
                if (r[i].x - 2 == x && (r[i].y + 1 == y || r[i].y - 1 == y))return false;
            if (map[r[i].x][r[i].y + 1] == 0)
                if ((r[i].x + 1 == x || r[i].x - 1 == x) && r[i].y + 2 == y)return false;
        }
    }
    return true;
}
bool jiangsi(int bx, int by)//将死没
{
    if (check(bx, by - 1)||check(bx, by + 1) || check(bx + 1, by) || check(bx - 1, by))
        return false;
    return true;
}
int main()
{
    int bx, by;
    while (cin >> n >> by >> bx && n != 0)
    {
        memset(map, 0, sizeof(map));
        for (int i = 1; i <= n; i++)
        {
            cin >> r[i].name >> r[i].y >> r[i].x;
            map[r[i].x][r[i].y] = r[i].name;
        }
        if (jiangsi(bx, by)) cout<<"YES\n";
        else cout<<"NO\n";
    }
    return 0;
}