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Gomoku

You are probably not familiar with the title, “Gomoku”, but you must have played it a lot. Gomoku is an abstract strategy board game and is also called Five in a Row, or GoBang. It is traditionally played with go pieces (black and white stones) on a go board (19x19 intersections). Nowadays, standard chessboard of Gomoku has 15x15 intersections. Black plays first, and players alternate in placing a stone of their color on an empty intersection. The winner is the first player to get an unbroken row of five or more stones horizontally, vertically, or diagonally. 

For convenience, we coordinate the chessboard as illustrated above. The left-bottom intersection is (0,0). And the bottom horizontal edge is x-axis, while the left vertical line is y-axis. 

I am a fan of this game, actually. However, I have to admit that I don’t have a sharp mind. So I need a computer program to help me. What I want is quite simple. Given a chess layout, I want to know whether someone can win within 3 moves, assuming both players are clever enough. Take the picture above for example. There are 31 stones on it already, 16 black ones and 15 white ones. Then we know it is white turn. The white player must place a white stone at (5,8). Otherwise, the black player will win next turn. After that, however, the white player also gets a perfect situation that no matter how his opponent moves, he will win at the 3rd move. 

So I want a program to do similar things for me. Given the number of stones and positions of them, the program should tell me whose turn it is, and what will happen within 3 moves.

31
3 3 1
3 4 0
3 5 0
3 6 0
4 4 1
4 5 1
4 7 0
5 3 0
5 4 0
5 5 1
5 6 1
5 7 1
5 9 1
6 4 1
6 5 1
6 6 0
6 7 1
6 8 0
6 9 0
7 5 1
7 6 0
7 7 1
7 8 1
7 9 0
8 5 0
8 6 1
8 7 0
8 8 1
8 9 0
9 7 1
10 8 0
1
7 7 1
1
7 7 0
0

Place white at (5,8) to win in 3 moves.
Cannot win in 3 moves.
Invalid.

2、判断对手是否存在两个空白格子使得他能够得到连续的5个棋子，因为这样，先手就不能堵住后手。

3、枚举任一空白格子放先手棋子，则只需对手方不存在“一空白格子使得棋子有连续的5个”，且先手方此时有"两个空白格子使得他能够得到连续的5个棋子",则先手胜。（攻防转换）

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define N 20
const int inf=0x3fffffff;
const double eps=1e-8;
int xx,yy;
int dx[4]={0,1,1,1};
int dy[4]={1,1,0,-1};
int g[N][N];
int bfs1(int v)  //判断是否存在一个空白格子使棋子连成连续的5个
{
    int i,j,k,x,y,sum;
    for(i=0;i<15;i++)
    {
        for(j=0;j<15;j++)
        {
            if(g[i][j]==-1)
            {
                for(k=0;k<4;k++)
                {
                    x=i+dx[k];
                    y=j+dy[k];
                    sum=1;
                    while(1)
                    {
                        if(x<0||x>=15||y<0||y>=15||g[x][y]!=v)
                            break;
                        sum++;
                        x+=dx[k];
                        y+=dy[k];
                    }
                    x=i-dx[k];
                    y=j-dy[k];
                    while(1)
                    {
                        if(x<0||x>=15||y<0||y>=15||g[x][y]!=v)
                            break;
                        sum++;
                        x-=dx[k];
                        y-=dy[k];
                    }
                    if(sum>=5)
                    {
                        xx=i;yy=j;
                        return 1;
                    }
                }
            }
        }
    }
    return 0;
}
int bfs2(int v) //判断是否存在2个空白格子使棋子连成连续的5个
{
    int i,j,k,x,y,sum,num=0;
    for(i=0;i<15;i++)
    {
        for(j=0;j<15;j++)
        {
            if(g[i][j]==-1)
            {
                for(k=0;k<4;k++)
                {
                    x=i+dx[k];
                    y=j+dy[k];
                    sum=1;
                    while(1)
                    {
                        if(x<0||x>=15||y<0||y>=15||g[x][y]!=v)
                            break;
                        sum++;
                        x+=dx[k];
                        y+=dy[k];
                    }
                    x=i-dx[k];
                    y=j-dy[k];
                    while(1)
                    {
                        if(x<0||x>=15||y<0||y>=15||g[x][y]!=v)
                            break;
                        sum++;
                        x-=dx[k];
                        y-=dy[k];
                    }
                    if(sum>=5)
                    {
                        if(num==1)
                            return 1;
                        num++;
                        break;
                    }
                }
            }
        }
    }
    return 0;
}
int bfs3(int v) //情况3,
{
    int i,j;
    for(i=0;i<15;i++)
    {
        for(j=0;j<15;j++)
        {
            if(g[i][j]==-1)
            {
                g[i][j]=v;
                if(bfs1(1-v)==0&&bfs2(v)==1)
                {
                    xx=i;yy=j;
                    return 1;
                }
                g[i][j]=-1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,n,x,y,d,w,b,val;
    while(scanf("%d",&n),n)
    {
        memset(g,-1,sizeof(g));
        w=b=0;
        for(i=0;i<n;i++)
        {
            scanf("%d%d%d",&x,&y,&d);
            g[x][y]=d;
            if(d==0)
                w++;
            else
                b++;
        }
        if(w>b)
        {
            printf("Invalid.\n");
            continue;
        }
        if(w==b)
            val=1;
        else
            val=0;
        if(bfs1(val))
        {
            if(val==0)
                printf("Place white at (%d,%d) to win in 1 move.\n",xx,yy);
            else
                printf("Place black at (%d,%d) to win in 1 move.\n",xx,yy);
        }
        else if(bfs2(1-val))
            printf("Lose in 2 moves.\n");
        else
        {
            if(bfs3(val))
            {
                if(val==0)
                    printf("Place white at (%d,%d) to win in 3 moves.\n",xx,yy);
                else
                    printf("Place black at (%d,%d) to win in 3 moves.\n",xx,yy);
            }
            else
                printf("Cannot win in 3 moves.\n");
        }
    }
    return 0;
}

hdu 3683 Gomoku （模拟、搜索）

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原文地址：http://blog.csdn.net/u011721440/article/details/40041431