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;

}

View Code

 

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;

}

View Code