poj 1324（BFS+状态压缩）

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解题思路：这道题一开始的想法就是状态压缩，即考虑如何判重，由于蛇并非是直线的，所以想到了以每一个点的上下左右共四个

值来表示相对位置。最开始想如何用四进制来表示它，无语。。。。。还是题目做少了，直接用两位来表示一个点即可（两位的二

进制数可以表示0-3）。剩下的关键就是判断蛇头会不会与蛇尾发生碰撞，详细的就看代码吧。。

整体的思路还是比较简单，但是代码很复杂，一般这种复杂点的搜索题代码都挺长的，所以很容易出错，不过可以用A*算法，而且

确实比朴素的算法要快，暂时还没想清楚怎么做。。。

#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(){    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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