1589象棋(大模拟)

题意：黑棋只有一个“将”，红棋有“马”，“炮”，“车”，“帅”中的几种，问黑棋是否没有被红棋将军。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#define repu(i,a,b) for(int i=a;i<b;i++)
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
char graph[12][12],ch[3];
int nx,ny;
int go[8][2] = {{1,0},{0,1},{-1,0},{0,-1},{1,1},{-1,-1},{1,-1},{-1,1}};
int inBlackPalace(int x,int y)
{
    if(x>=1&&x<=3&&y>=4&&y<=6)
        return 1;
    return 0;
}
int goG(int x,int y)
{
    int a,b;
    if(y==ny)
    {
        if(x>nx)
            a=nx,b=x;
        else
            a=x,b=nx;
        repu(i,a,b)
        if(graph[i][y])///帅与将之间有间隔
            return 0;
        return 1;///被帅将军
    }
    return 0;
}
int goH(int x,int y)///蹩马腿
{
    ///与马能构成“日”字
    if(abs(nx-x)==2&&abs(ny-y)==1)
    {
        if(graph[(nx+x)/2][y])
            return 0;
        return 2;///被马将军
    }
    if(abs(ny-y)==2&&abs(nx-x)==1)
    {
        if(graph[x][(ny+y)/2])
            return 0;
        return 2;///被马将军
    }
    return 0;
}
int goR(int x,int y)
{
    int a,b;
    if(x==nx)
    {
        if(y>ny)
            a=ny,b=y;
        else
            a=y,b=ny;
        repu(i,a,b)
        if(graph[x][i])
            return 0;
        return 3;///被车将军
    }
    if(y==ny)
    {
        if(x>nx)
            a=nx,b=x;
        else
            a=x,b=nx;
        repu(i,a,b)
        if(graph[i][y])
            return 0;
        return 3;///被车将军
    }
    return 0;
}
int goC(int x,int y)
{
    int sum = 0,a,b;
    if(x==nx&&y>ny)
    {
        repu(i,y,ny)
        {
            if(graph[x][i])
                sum++;
        }
        if(sum==1)
            return 4;///被炮将军
        return 0;
    }
    if(y==ny)
    {
        if(x>nx)
            a=nx,b=x;
        else
            a=x,b=nx;
        repu(i,a,b)
        {
            if(graph[i][y])
                sum++;
        }
        if(sum==1)
            return 4;
        return 0;
    }
    return 0;
}
int goo(int x, int y)
{
    //  printf("%d-%d\n",nx,ny);
    switch(graph[x][y])
    {
    case ‘G‘:
        return goG(x,y);
    case ‘R‘:
        return goR(x,y);
    case ‘H‘:
        return goH(x,y);
    case ‘C‘:
        return goC(x,y);
    }
    return false;
}
int main()
{
    int n;
    while (scanf("%d", &n) != EOF)
    {
        int bx, by,temp = false,o = 0;
        scanf("%d%d", &bx, &by);
        if (n == 0 && bx == 0 && by == 0) break;

        memset(graph, 0, sizeof(graph));

        for (int i = 0; i < n; i++)
        {
            int x, y;
            scanf("%s%d%d", ch, &x, &y);
            graph[x][y] = ch[0];
        }
        if((bx==1&&by==4)||(bx==1&&by==6)||(bx==3&&by==4)||(bx==3&&by==6))
            o = 8;
        else
            o = 4;
        for (int way = 0; way < o; way++)
        {
            nx = bx + go[way][0];
            ny = by + go[way][1];
            if (!inBlackPalace(nx, ny))
                continue;
            char tep = graph[nx][ny];
            graph[nx][ny] = 0;
            //  printf("way is--%d\n",way);
            int ok = 1;
            for (int i = 1; i <= 10; i++)
            {
                for (int j = 1; j <= 9; j++)
                {
                    if(graph[i][j])
                    {
                        // printf("%d-%d:%c\n",i,j,graph[i][j]);
                        if(goo(i,j))///将军成功了,再换条路
                        {
                            ok =0;
                            break;
                        }
                    }
                }
                if(ok==0)
                    break;
            }
            //  printf("****\n");
            graph[nx][ny] = tep;
            if(ok)///此路可以不被将军
            {
                temp = true;
                break;
            }
        }
        if(temp)
            puts("NO");
        else
            puts("YES");
    }
    return 0;
}