COJ 1046 追杀

马可以从任意位置出发，走遍整个棋盘；

先用 bfs 求出马到达每个位置的最短时间 Ti，然后模拟将的移动，当将移动的时间 Tk 满足 Tk>=Ti 且Tk-Ti为偶数时相遇（马可以在两个位置徘徊一会等待将的到来）；

 

# include <stdio.h>

# include <string.h>

const int dir[][2] = {{1,2}, {2,1}, {-1,2}, {1,-2}, {-2,1}, {2,-1}, {-1,-2}, {-2,-1}};

int nx, ny, kx, ky;

char vis[9][8], dis[9][8];

void bfs(void)

{

    char Q[85][2];

    int front, rear, i, cx, cy, x, y;

    memset(vis, 0, sizeof(vis));

    vis[nx][ny] = 1;

    dis[nx][ny] = 0;

    Q[1][0] = nx, Q[1][1] = ny;

    front = 1, rear = 2;

    while(front < rear)

    {

        cx = Q[front][0], cy = Q[front][1];

        ++front;

        for( i = 0; i < 8; ++i )

        {

            x = cx + dir[i][0], y = cy + dir[i][1];

            if(0<=x&&x<=8 && 0<=y&&y<=7 && !vis[x][y])

            {

                Q[rear][0] = x, Q[rear][1] = y;

                ++rear;

                dis[x][y] = dis[cx][cy] + 1;

                vis[x][y] = 1;

            }

        }

    }

}



int search(void)

{

    int t, dx, dy;



    t = 0;

    dx = 1, dy = 1;

    while(1)

    {

        if(t>=dis[kx][ky] && (t-dis[kx][ky])%2==0) return t;

        ++t;

        if(0==kx && -1==dx) dx = 1;  // 向右

        if(8==kx && 1==dx) dx = -1;  // 向左

        if(0==ky && -1==dy) dy = 1;  // 向下

        if(7==ky && 1==dy) dy = -1;  // 向上

        kx += dx, ky += dy;

    }

}



int main()

{

    while(~scanf("%d%d%d%d", &nx,&ny,&kx,&ky))

    {

        bfs();

        printf("%d\n", search());

    }

}

//