练习题 贪吃蛇走迷宫 POJ1324

/*POJ 1324

题意：

给出一个n*m的贪吃蛇地图，以及贪吃蛇现在身体各个块所在的位置，求它的头走到（1,1）位置最少需要多少步。

其中蛇在移动的过程中不能走到障碍物上，也不能撞到自己的身体上。

题解：写迷宫问题首先应该避免死循环，即相同状态不能多次入队。本次搜索状态不仅有蛇头的位置还有蛇身的相对位置。

因为蛇身必然连贯，利用这点进行状态压缩，相对于上个点可以有4种方位。

可优化的地方：

由于vis数组很大,如果对于每个tc都初始化vis数组浪费时间,所以在申请了全局变量vis之后,我们对于每个tc,都只用当前的tc值去标记vis表示当前状态已出现。

*/

 

#include <stdio.h>
#define MAX 22
bool map[MAX][MAX],vis[MAX][MAX][16385];
struct Node{
    int x, y, st, d;
};
int bfs(int hy,int hx,int hst){
    if (hy == 1 && hx == 1)return 0;
    for (int i = 0; i < MAX;++i)
    for (int j = 0; j < MAX;++j)
    for (int k = 0; k < 16384; ++k)
        vis[i][j][k] = 0;
    q.init();
    q.push(hy, hx, hst, 0);
    vis[hy][hx][hst] = true;
    while (!q.empty()){
        Node in = q.pop();
        int y = in.y, x = in.x, st = in.st, d = in.d;
        for (int dn = 0; dn < 4; ++dn){
            int ny = y + dy[dn], nx = x + dx[dn];
            if (ny == 1 && nx == 1)return d + 1;
            if (ny<1 || ny>N || nx<1 || nx>M || map[ny][nx] || isSnake(ny, nx, in))continue;
            int nst = (st >> 2) + (((dn + 2) % 4) << (2 * (L - 2)));
            if (vis[ny][nx][nst])continue;
            q.push(ny, nx, nst, d + 1);
            vis[ny][nx][nst] = true;
        }
    }
    return -1;
}
int main(){
    int tc = 0, heady, headx, headst;
    while (scanf("%d%d%d", &N, &M, &L) != EOF && N){
        headst = 0;
        scanf("%d%d", &heady, &headx);
        int ty, tx, prey = heady, prex = headx;
        for (int i = 1; i < L; ++i){
            scanf("%d%d", &ty, &tx);
            for (int dn = 0; dn < 4; ++dn){
                if (prey + dy[dn] == ty&&prex + dx[dn] == tx){
                   headst = (headst << 2) + dn;
                   break;
                }
            }
            prey = ty, prex = tx;
        }  
        scanf("%d", &K);
        int block_x, block_y;
        for (block_x = 0; block_x < MAX; ++block_x)
        for (block_y = 0; block_y < MAX; ++block_y)
            map[block_x][block_y] = false;
        for (int i = 0; i < K; ++i){
            scanf("%d%d", &block_x, &block_y);
            map[block_x][block_y] = true;
        }
        printf("Case %d: %d\n", ++tc, bfs(heady, headx, headst));
    }
    return 0;
}
 
struct Queue{
    Node nodepool[1200000];//3^6*4*20*20 / 4^7*400
    Node *tnode;
    int poolIndex, front, rear;
    void init(){
        poolIndex = -1;
        front = rear = 0;
    }
    void push(int y, int x, int st, int d){
        tnode = &nodepool[rear++];
        tnode->y = y, tnode->x = x, tnode->st = st, tnode->d = d;
    }
    Node pop(){
        Node ret = nodepool[front++];
        return ret;
    }
    bool empty(){
        return rear == front;
    }
}q;
 
int dy[] = { 1, 0, -1, 0 };//下,左,上,右
int dx[] = { 0, -1, 0, 1 };
 
bool isSnake(int findy, int findx, Node nd){
    int dir[10];
    for (int i = L - 1; i > 0; --i){
        dir[i] = nd.st & 3;//取低两位
        nd.st >>= 2;
    }
    int y = nd.y, x = nd.x;
    for (int i = 1; i < L; ++i){
        y += dy[dir[i]];
        x += dx[dir[i]];
        if (y == findy&&x == findx)return true;
    }
    return false;
}