Poj3322 Bloxorz I

Poj3322 Bloxorz I

Classic box games

The minimum number of steps for a 1*2*1 brick reaches a specified hole.

Obviously bfs, the State indicates that a P value 0, 1, and 2 respectively indicate that the bricks are standing, lying horizontally and lying vertically, when weighting, use a three-dimensional array to obtain vis [p state] [row position] [column position].

Then, you can directly move from one state to another, and change the coordinates at the same time.

#include 
 
  #include 
  
   #include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include using namespace std;//LOOP#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))//OTHER#define PB push_back//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RS(s) scanf("%s", s)typedef long long LL;typedef unsigned long long ULL;typedef vector 
         
           VI;const double eps = 1e-9;const int MOD = 1000000007;const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const int maxn = 110;int n, m;char s[510][510];int d[510][510][3];bool vis[510][510][3];int dir[][4][3] = {{{-2, 0, 2}, {0, 1, 1}, {1, 0, 2}, {0, -2, 1}}, {{-1, 0, 0}, {0, 2, -1}, {1, 0, 0}, {0, -1, -1}}, {{-1, 0, -2}, {0, 1, 0}, {2, 0, -2}, {0, -1, 0}} };struct Node{ int x, y; int p; Node(){} Node(int x, int y, int p): x(x), y(y), p(p){} bool operator == (const Node& a) const { return x == a.x && y == a.y && p == a.p; }}st, ed, now, nxt;inline void getorder(int &x1, int &y1, int &x2, int &y2, int &p){ if (x1 > x2 || (x1 == x2 && y1 > y2)) { swap(x1, x2); swap(y1, y2); } if (x1 == x2) p = 1; else p = 2; if (x1 == x2 && y1 == y2) p = 0;}vector
          
            > stv;bool ok(int x, int y, int p){ if (x < 0 || x >= n || y >= m || y < 0) return 0; if (p == 0) { if (s[x][y] == '#' || s[x][y] == 'E') return 0; return 1; } if (s[x][y] == '#') return 0; int x2, y2; if (p == 1) x2 = x, y2 = y + 1; if (p == 2) x2 = x + 1, y2 = y; if (x2 < 0 || x2 >= n || y2 >= m || y2 < 0) return 0; if (s[x2][y2] == '#') return 0; return 1;}int bfs(){ if (st == ed) return 0; queue
           
            q; int x, y, p; CLR(vis, 0); d[st.x][st.y][st.p] = 0; vis[st.x][st.y][st.p] = 1; q.push(st); while (!q.empty()) { now = q.front(); q.pop(); REP(i, 4) { x = now.x + dir[now.p][i][0]; y = now.y + dir[now.p][i][1]; p = now.p + dir[now.p][i][2]; if (!ok(x, y, p) || vis[x][y][p]) continue; nxt.x = x, nxt.y = y, nxt.p = p; if (nxt == ed) return d[now.x][now.y][now.p] + 1; d[x][y][p] = d[now.x][now.y][now.p] + 1; vis[x][y][p] = 1; q.push(nxt); } } return -1;}int main(){ //freopen("0.txt", "r", stdin); while (~RII(n, m) && n + m) { stv.clear(); REP(i, n) { RS(s[i]); REP(j, m) { if (s[i][j] == 'X') { s[i][j] = '.'; stv.push_back(make_pair(i, j)); } if (s[i][j] == 'O') { s[i][j] = '.'; ed.x = i; ed.y = j; ed.p = 0; } } } int pp = 0; if (stv.size() == 2) getorder(stv[0].first, stv[0].second, stv[1].first, stv[1].second, pp); st.x = stv[0].first; st.y = stv[0].second; st.p = pp; int ans = bfs(); if (ans == -1) puts("Impossible"); else printf("%d\n", ans); } return 0;}