codevs1004 四子连棋

题目描述 Description

在一个4*4的棋盘上摆放了14颗棋子，其中有7颗白色棋子，7颗黑色棋子，有两个空白地带，任何一颗黑白棋子都可以向上下左右四个方向移动到相邻的空格，这叫行棋一步，黑白双方交替走棋，任意一方可以先走，如果某个时刻使得任意一种颜色的棋子形成四个一线（包括斜线），这样的状态为目标棋局。

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输入描述 Input Description

从文件中读入一个4*4的初始棋局，黑棋子用B表示，白棋子用W表示，空格地带用O表示。

输出描述 Output Description

用最少的步数移动到目标棋局的步数。

样例输入 Sample Input

BWBO
WBWB
BWBW
WBWO

样例输出 Sample Output

5

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>

using namespace std;
const int maxn = ;

struct member{
    int last,step;
    int board[][];//[y][x]
};
member first,q[maxn];
int tot;
int dx[] = {,-,,};
int dy[] = {-,,,};

bool judge(member temp){
    int r1 = ,r2 = ;
    for(r1 = ;r1 <= ;r1++){
        bool sign = true;
        for(r2 = ;r2 <= ;r2++){
            if(temp.board[r1][r2] != temp.board[r1][r2+]) sign = false;
        }
        if(sign) return true;
    }
    for(r1 = ;r1 <= ;r1++){
        bool sign = true;
        for(r2 = ;r2 <= ;r2++){
            if(temp.board[r2][r1] != temp.board[r2+][r1]) sign = false;
        }
        if(sign) return true;
    }
    if(temp.board[][] == temp.board[][] && temp.board[][] == temp.board[][] && temp.board[][] == temp.board[][]) return true;
    if(temp.board[][] == temp.board[][] && temp.board[][] == temp.board[][] && temp.board[][] == temp.board[][]) return true;
    return false;
}

int bfs(){
    int h = ,t = ,nx,ny;
    member tailer;
    while(h <= t){
        if(judge(q[h])) return q[h].step;
        int f1 = ,f2 = ,ya,xa,yb,xb,sign = ;
        for(f1 = ;f1 <= ;f1++){
            for(f2 = ;f2 <= ;f2++){
                if(q[h].board[f1][f2] == ){
                    if(sign == ){
                        sign = ;
                        ya = f1;
                        xa = f2;
                    }else if(sign == ){
                        yb = f1;
                        xb = f2;
                    }
                }
            }
        }
        f1 = f2 = ;
        for(f1 = ;f1 < ;f1++){
            nx = xa + dx[f1];
            ny = ya + dy[f1];
            if(nx <=  && nx >=  && ny <= && ny >=  && q[h].board[ny][nx] != q[h].last){
                t++;
                tailer = q[h];
                tailer.last = tailer.board[ny][nx];
                swap(tailer.board[ny][nx],tailer.board[ny - dy[f1]][nx - dx[f1]]);
                tailer.step++;
                q[t] = tailer;

            }
            nx = xb + dx[f1];
            ny = yb + dy[f1];
            if(nx <=  && nx >=  && ny <= && ny >=  && q[h].board[ny][nx] != q[h].last){
                t++;
                tailer = q[h];
                tailer.last = tailer.board[ny][nx];
                swap(tailer.board[ny][nx],tailer.board[ny - dy[f1]][nx - dx[f1]]);
                tailer.step++;
                q[t] = tailer;

            }

        }
        h++;
    }

}

int main(){
    char cmd;
    int r1,r2,ans;
    for(r1 = ;r1 <= ;r1++){
        for(r2 = ;r2 <= ;r2++){
            cin>>cmd;
            if(cmd == 'O') first.board[r1][r2] = ;
            if(cmd == 'B') first.board[r1][r2] = ;
            if(cmd == 'W') first.board[r1][r2] = ;
        }
    }
    first.step = ;
    first.last = ;
    q[] = first;
    ans = bfs();
    cout<<ans;
    return ;
}