POJ 1324 Holedox Moving(状态压缩BFS)

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题目链接：POJ 1324 Holedox Moving

状态压缩比较容易看出来，但是怎么实现我就不会了，看了人家题解又写了好久才搞出来。。得练练状压了。

用两个二进制位表示一块蛇身相对于前一块的方向，那么最多14位就可以表示蛇身，仔细观察能看出来，随着蛇头移动，后一个相对位置和移动前的前一个的相对位置是一样的，蛇头后边那个与蛇头与移动方向的相对方向一样，那么就可以使用二进制的移位操作实现了。

方向数组的顺序不能乱写，与judge_code的dir是一一对应的。

#include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;const int MAX_S = (1 << 14) + 100;const int MAX_N = 20 + 2;const int INF = (1 << 29);struct State{    int x, y, dis, s;    State(int x = 0, int y = 0, int dis = 0, int s = 0) : x(x), y(y), dis(dis), s(s) {};};struct Point{    int x, y;};int N, M, res, L;int vis[MAX_N][MAX_N][MAX_S];int fx[4] = {-1, 0, 1, 0};int fy[4] = {0, 1, 0, -1};bool _map[MAX_N][MAX_N];Point pos[MAX_N * MAX_N];queue <State> Q;int get_start(){    int dir, dx, dy, s = 0;    for(int i = L - 1; i > 0; i--)    {        dx = pos[i].x - pos[i - 1].x, dy = pos[i].y - pos[i - 1].y;        if(dx == 0 && dy == 1)            dir = 1;        else if(dx == 0 && dy == -1)            dir = 3;        else if(dx == -1 && dy == 0)            dir = 0;        else if(dx == 1 && dy == 0)            dir = 2;        s = s << 2;        s = s | dir;    }    return s;}int get_next_state(int i, int s){    int dir;    int k = (1 << ((L - 1) << 1)) - 1;    int dx = 0, dy = 0;    dx = dx - fx[i], dy = dy - fy[i];    if(dx == 0 && dy == 1)        dir = 1;    else if(dx == 0 && dy == -1)        dir = 3;    else if(dx == -1 && dy == 0)        dir = 0;    else if(dx == 1 && dy == 0)        dir = 2;    s = s << 2;    s = s | dir;    s = s & k; // 去除高位部分    return s;}bool judge_code(int x, int y, int pre_x, int pre_y, int s){    int dir;    for(int i = 0; i < L - 1; i++)    {        dir = 3;        dir = dir & s;        s = s >> 2;        if(x == pre_x + fx[dir] && y == pre_y + fy[dir])            return false;        pre_x = pre_x + fx[dir], pre_y = pre_y + fy[dir];    }    return true;}void BFS(){    State a;    int dx, dy, s;    while(!Q.empty())    {        a = Q.front();        Q.pop();        for(int i = 0; i < 4; i++)        {            dx = a.x + fx[i], dy = a.y + fy[i];            s = get_next_state(i, a.s);            if(dx > 0 && dy > 0 && dx <= N && dy <= M && !vis[dx][dy][s] && !_map[dx][dy] && judge_code(dx, dy, a.x, a.y, a.s))            {                if(dx == 1 && dy == 1)                {                    res = a.dis + 1;                    return ;                }                vis[dx][dy][s] = 1;                Q.push(State(dx, dy, a.dis + 1, s));            }        }    }}int main(){    //freopen("in.txt", "r", stdin);    int s = 0, _case = 0;    State _start;    while(scanf("%d%d%d", &N, &M, &L), N + M + L)    {        res = INF;        memset(_map, 0 , sizeof(_map));        memset(vis, 0 , sizeof(vis));        for(int i = 0; i < L; i++)            scanf("%d%d", &pos[i].x, &pos[i].y);        int K, u, v;        scanf("%d", &K);        for(int i = 0; i < K; i++)        {            scanf("%d%d", &u, &v);            _map[u][v] = 1;        }        if(pos[0].x == 1 && pos[0].y == 1)        {            printf("Case %d: 0\n", ++_case);            continue;        }        s = get_start();        Q.push(State(pos[0].x, pos[0].y, 0, s));        vis[pos[0].x][pos[0].y][s] = 1;        BFS();        if(res == INF)            printf("Case %d: -1\n", ++_case);        else            printf("Case %d: %d\n", ++_case, res);        while(!Q.empty())            Q.pop();    }    return 0;}

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