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());
    }
}

//