POJ 3322 Bloxorz I BFS

一道有趣的搜索题，用广搜比较好。注意记录状态即可， 然后更新好标记数组。每次是根据方块左上角的坐标来进行操作的。

判断能不能走的时候一定要注意占两个格的情况，到达终点时一定是竖起来才行，薄弱地带一定不能竖起来。

队列可以用STL 也可以用数组，不过用数组的话，交到G++上，可以优化到700ms或者更少

/* ID: sdj22251 PROG: subset LANG: C++ */ #include <iostream> #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <cmath> #include <ctime> #define LOCA using namespace std; int n, m; char s[505][505]; bool v[505][505][5]; int move[4][2][4] = {{{0, 0, 0, 0},{0, 0, 0, 0}}, {{0, 0, -2, 1},{1, -2, 0, 0}}, {{-1, 1, 0, 0},{0, 0, -1, 2}}, {{0, 0, -1, 2},{-1, 1, 0, 0}} }; struct wwj { int x, y, con, num; wwj() {} wwj(int a, int b, int c, int d) { x = a; y = b; con = c; num = d; } }q[505* 505 * 3]; bool check(wwj x) { if(x.x >= n || x.y >= m || x.x < 0 || x.y < 0) return false; if(x.con == 1) { if(s[x.x][x.y] == '#') return false; } else if(x.con == 2) { if(x.y + 1 >= m || s[x.x][x.y + 1] == '#' || s[x.x][x.y] == '#') return false; } else if(x.con == 3) { if(x.x + 1 >= n || s[x.x][x.y] == '#' || s[x.x + 1][x.y] == '#') return false; } return true; } int main() { #ifdef LOCAL freopen("subset.in","r",stdin); freopen("subset.out","w",stdout); #endif int i, j, start_row, end_row, start_clu, end_clu, con; while(scanf("%d%d", &n, &m) != EOF) { if(n == 0 && m == 0) break; memset(v, 0, sizeof(v)); con = 0; for(i = 0; i < n; i++) scanf("%s", s[i]); for(i = 0; i < n; i++) { for(j = 0; j < m; j++) { if(s[i][j] == 'X') { if(!con) { start_row = i; start_clu = j; con = 1; } else if(start_clu + 1 == j) con = 2; else con = 3; } if(s[i][j] == 'O') { end_row = i; end_clu = j; } } } int head = 0, tail = 0; wwj A(start_row, start_clu, con, 0); q[tail++] = A; v[A.x][A.y][A.con] = true; int ans = -1; while(head < tail) { A = q[head++]; if(A.x == end_row && A.y == end_clu && A.con == 1) { ans = A.num; break; } for(i = 0; i < 4; i++) { int tx = A.x + move[A.con][0][i]; int ty = A.y + move[A.con][1][i]; int tcon = A.con; int tnum = A.num + 1; if(tcon == 1 && i < 2) tcon = 2; else if(tcon == 1 && i >= 2) tcon = 3; else if(tcon == 2 && i >= 2) tcon = 1; else if(tcon == 3 && i >= 2) tcon = 1; wwj B(tx, ty, tcon, tnum); if(check(B) && !v[tx][ty][tcon]) { if(s[tx][ty] == 'E') { if(tcon != 1) { q[tail++] = B; v[B.x][B.y][B.con] = true; } } else { q[tail++] = B; v[B.x][B.y][B.con] = true; } } } } if(ans != -1) printf("%d\n", ans); else puts("Impossible"); } return 0; }