HDU 4121 Xiangqi --模拟

题意： 给一个象棋局势，问黑棋是否死棋了，黑棋只有一个将，红棋可能有2~7个棋，分别可能是车，马，炮以及帅。

解法： 开始写法是对每个棋子，都处理处他能吃的地方，赋为-1，然后判断将能不能走到非-1的点。但是WA了好久，也找不出反例，但就是觉得不行，因为可能有将吃子的情况，可能有hack点。但是比赛后还是被我调出来了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;
#define N 1017

int chess[14][13],mp[14][13];  //0 : Non  1:G帅 2:R车 3:H Horse 4:C Cannon
int dx[4] = {0,-1,0,1};
int dy[4] = {1,0,-1,0};
bool InPalace(int nx,int ny) { if(nx >= 1 && nx <= 3 && ny >= 4 && ny <= 6) return true; return false; }
bool InChess(int nx,int ny)  { if(nx >= 1 && nx <= 10 && ny >= 1 && ny <= 9) return true; return false;}

int main()
{
    int n,X,Y,i,j,k;
    int x,y;
    char ss[4];
    while(scanf("%d%d%d",&n,&X,&Y)!=EOF && (n+X+Y))
    {
        memset(chess,0,sizeof(chess));
        memset(mp,0,sizeof(mp));
        for(i=1;i<=n;i++)
        {
            //0 : Non  1:G帅 2:R车 3:H Horse 4:C Cannon
            scanf("%s%d%d",ss,&x,&y);
            if(ss[0] == 'G')      chess[x][y] = 1;
            else if(ss[0] == 'R') chess[x][y] = 2;
            else if(ss[0] == 'H') chess[x][y] = 3;
            else if(ss[0] == 'C') chess[x][y] = 4;
        }
        for(i=1;i<=10;i++)
        {
            for(j=1;j<=9;j++)
            {
                if(chess[i][j] == 1) // shuai
                {
                    for(k=i-1;k>=1;k--)
                    {
                        if(chess[k][j] != 0)
                        {
                            mp[k][j] = -1;
                            break;
                        }
                        if(k <= 3 && chess[k][j] == 0) mp[k][j] = -1;
                    }
                }
                else if(chess[i][j] == 2)  // R che
                {
                    for(int D=0;D<4;D++)
                    {
                        int kx = i, ky = j;
                        while(1)
                        {
                            kx = kx + dx[D];
                            ky = ky + dy[D];
                            if(!InChess(kx,ky)) break;
                            if(chess[kx][ky] == 0) mp[kx][ky] = -1;
                            else
                            {
                                mp[kx][ky] = -1;
                                break;
                            }
                        }
                    }
                }
                else if(chess[i][j] == 3)   //Horse
                {
                    if(InChess(i-1,j) && chess[i-1][j] == 0 && (i-1 != X || j != Y))  //UP not blocked
                    {
                        if(InChess(i-2,j-1)) mp[i-2][j-1] = -1;
                        if(InChess(i-2,j+1)) mp[i-2][j+1] = -1;
                    }
                    if(InChess(i+1,j) && chess[i+1][j] == 0 && (i+1 != X || j != Y))  //DOWN not blocked
                    {
                        if(InChess(i+2,j-1)) mp[i+2][j-1] = -1;
                        if(InChess(i+2,j+1)) mp[i+2][j+1] = -1;
                    }
                    if(InChess(i,j+1) && chess[i][j+1] == 0 && (i != X || j+1 != Y))  //RIGHT not blocked
                    {
                        if(InChess(i-1,j+2)) mp[i-1][j+2] = -1;
                        if(InChess(i+1,j+2)) mp[i+1][j+2] = -1;
                    }
                    if(InChess(i,j-1) && chess[i][j-1] == 0 && (i != X || j-1 != Y))  //LEFT not blocked
                    {
                        if(InChess(i-1,j-2)) mp[i-1][j-2] = -1;
                        if(InChess(i+1,j-2)) mp[i+1][j-2] = -1;
                    }
                }
                else if(chess[i][j] == 4)   //Cannon pao
                {
                    for(int D=0;D<4;D++)
                    {
                        int kx = i, ky = j;
                        int cnt = 0;
                        while(1)
                        {
                            kx = kx + dx[D];
                            ky = ky + dy[D];
                            if(!InChess(kx,ky)) break;
                            if(cnt == 1 && chess[kx][ky] == 0) mp[kx][ky] = -1;
                            if(chess[kx][ky] != 0)
                            {
                                if(cnt == 0) cnt++;
                                else if(cnt == 1)
                                {
                                    mp[kx][ky] = -1;
                                    break;
                                }
                            }
                        }
                    }
                }
            }
        }
        int tag = 0;
        for(k=0;k<4;k++)
        {
            int kx = X + dx[k];
            int ky = Y + dy[k];
            if(!InChess(kx,ky) || !InPalace(kx,ky)) continue;
            if(mp[kx][ky] != -1) { tag = 1; break; }  //可以走
        }
        if(tag) puts("NO");
        else    puts("YES");
    }
    return 0;
}

比赛中后来换了一种写法，枚举将能走的四个位置，然后判断此时是否能被吃掉。 如果没有一个安全的地方，那么就死棋了。 这种写起来就好些多了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
#define N 1017

int chess[14][13],mp[14][13];  //0 : Non  1:G帅 2:R车 3:H Horse 4:C Cannon
int dx[4] = {0,-1,0,1};
int dy[4] = {1,0,-1,0};

bool InPalace(int nx,int ny)
{
    if(nx >= 1 && nx <= 3 && ny >= 4 && ny <= 6) return true;
    return false;
}

bool InChess(int nx,int ny)
{
    if(nx >= 1 && nx <= 10 && ny >= 1 && ny <= 9) return true;
    return false;
}

struct node{
    int x,y,type;
}q[10];

bool Check(int X,int Y,int i,int j,int type)  //将死则 return false !
{
    int k;
    if(type == 1)     //帅
    {
        if(j != Y) return true;   //不是一行
        for(k=i-1;k>X;k--)
        {
            if(chess[k][j] != 0)
                break;
        }
        if(k == X) return false;  //老倌子见面,gg
        return true;
    }
    else if(type == 2)       //R
    {
        int tag = 1;
        for(int D=0;D<4;D++)
        {
            int kx = i, ky = j;
            while(1)
            {
                kx = kx + dx[D];
                ky = ky + dy[D];
                if(!InChess(kx,ky)) break;
                if(kx == X && ky == Y)  //车能碰到将
                {
                    tag = 0;
                    break;
                }
                if(chess[kx][ky] != 0)
                    break;
            }
        }
        if(!tag) return false;
        return true;
    }
    else if(type == 3)  //Horse
    {
        int tag = 1;
        if(InChess(i-1,j) && chess[i-1][j] == 0 && (i-1 != X && j != Y))  //UP not blocked
        {
            if(InChess(i-2,j-1) && i-2 == X && j-1 == Y)
                tag = 0;
             if(InChess(i-2,j+1) && i-2 == X && j+1 == Y)
                tag = 0;
        }
        if(InChess(i+1,j) && chess[i+1][j] == 0 && (i+1 != X && j != Y))  //DOWN not blocked
        {
            if(InChess(i+2,j-1) && i+2 == X && j-1 == Y)
                tag = 0;
            if(InChess(i+2,j+1) && i+2 == X && j+1 == Y)
                tag = 0;
        }
        if(InChess(i,j+1) && chess[i][j+1] == 0 && (i != X && j+1 != Y))  //RIGHT not blocked
        {
            if(InChess(i-1,j+2) && i-1 == X && j+2 == Y)
                tag = 0;
            if(InChess(i+1,j+2) && i+1 == X && j+2 == Y)
                tag = 0;
        }
        if(InChess(i,j-1) && chess[i][j-1] == 0 && (i != X && j-1 != Y))  //LEFT not blocked
        {
            if(InChess(i-1,j-2) && i-1 == X && j-2 == Y)
                tag = 0;
            if(InChess(i+1,j-2) && i+1 == X && j-2 == Y)
                tag = 0;
        }
        if(!tag) return false;
        return true;
    }
    else if(type == 4)         //Cannon
    {
        int tag = 1;
        for(int D=0;D<4;D++)
        {
            int kx = i, ky = j;
            int cnt = 0;
            while(1)
            {
                kx = kx + dx[D];
                ky = ky + dy[D];
                if(!InChess(kx,ky)) break;
                if(cnt == 1 && kx == X && ky == Y)
                {
                    tag = 0;
                    break;
                }
                if(chess[kx][ky] != 0)
                    cnt++;
            }
        }
        if(!tag) return false;
        return true;
    }
    return false;
}

int main()
{
    int n,X,Y,i,j,k;
    int x,y;
    char ss[4];
    while(scanf("%d%d%d",&n,&X,&Y)!=EOF && (n+X+Y))
    {
        memset(chess,0,sizeof(chess));
        int tot = 0;
        for(i=1;i<=n;i++)
        {
            //0 : Non  1:G帅 2:R车 3:H Horse 4:C Cannon
            scanf("%s %d%d",ss,&x,&y);
            if(ss[0] == 'G')
                chess[x][y] = 1;
            else if(ss[0] == 'R')
                chess[x][y] = 2;
            else if(ss[0] == 'H')
                chess[x][y] = 3;
            else if(ss[0] == 'C')
                chess[x][y] = 4;
            node now;
            now.x = x, now.y = y, now.type = chess[x][y];
            q[++tot] = now;
        }
        int flag[10];
        int tag = 0;
        for(k=0;k<4;k++)
        {
            int nx = X + dx[k];
            int ny = Y + dy[k];
            memset(flag,0,sizeof(flag));
            if(!InChess(nx,ny) || !InPalace(nx,ny)) continue;
            for(i=1;i<=tot;i++)
            {
                if(nx == q[i].x && ny == q[i].y)
                {
                    flag[i] = 1;
                    chess[nx][ny] = 0;
                }
            }
            int smalltag = 1;
            for(i=1;i<=tot;i++)
            {
                if(flag[i]) continue;
                bool res = Check(nx,ny,q[i].x,q[i].y,q[i].type);
                if(!res) { smalltag = 0; break; }
            }
            if(smalltag)
            {
                tag = 1;
                break;
            }
            // 还原
            for(i=1;i<=tot;i++)
            {
                if(flag[i])
                    chess[q[i].x][q[i].y] = q[i].type;
            }
        }
        if(tag) puts("NO");
        else    puts("YES");
    }
    return 0;
}