POJ 3322 Bloxorz(算競進階習題)
阿新 • • 發佈:2019-03-16
end out printf sca ini 狀態 ems turn inline
bfs
標準廣搜題,主要是把每一步可能的坐標都先預處理出來,會好寫很多
每個狀態對應三個限制條件,x坐標、y坐標、lie=0表示直立在(x,y),lie=1表示橫著躺,左半邊在(x,y),lie=2表示豎著躺,上半邊在(x,y)
#include <bits/stdc++.h> #define INF 0x3f3f3f3f using namespace std; typedef long long ll; inline int lowbit(int x){ return x & (-x); } inline int read(){ int X = 0, w = 0; char ch = 0; while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); } while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar(); return w ? -X : X; } inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; } inline int lcm(int a, int b){ return a / gcd(a, b) * b; } template<typename T> inline T max(T x, T y, T z){ return max(max(x, y), z); } template<typename T> inline T min(T x, T y, T z){ return min(min(x, y), z); } template<typename A, typename B, typename C> inline A fpow(A x, B p, C lyd){ A ans = 1; for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd; return ans; } char g[505][505]; struct rec { int x, y, lie; } st, ed; int r, c; const int dx[] = {0, 0, 1, -1}; const int dy[] = {1, -1, 0, 0}; int f[505][505][3]; const int nextx[3][4] = {{-2, 1, 0, 0}, {-1, 1, 0, 0}, {-1, 2, 0, 0}}; const int nexty[3][4] = {{0, 0, -2, 1}, {0, 0, -1, 2}, {0, 0, -1, 1}}; const int nextlie[3][4] = {{2, 2, 1, 1}, {1, 1, 0, 0}, {0, 0, 2, 2}}; bool inArea(int x, int y){ return x >= 0 && x < r && y >= 0 && y < c; } bool valid(const rec &next){ int x = next.x, y = next.y, lie = next.lie; if(!inArea(x, y)) return false; if(g[x][y] == '#') return false; if(lie == 0 && g[x][y] == 'E') return false; if(lie == 1 && g[x][y + 1] == '#') return false; if(lie == 2 && g[x + 1][y] == '#') return false; return true; } void init(){ for(int i = 0; i < r; i ++){ for(int j = 0; j < c; j ++){ if(g[i][j] == 'O') ed.x = i, ed.y = j, ed.lie = 0, g[i][j] = '.'; if(g[i][j] == 'X'){ bool flag = false; for(int d = 0; d < 4; d ++){ int nx = i + dx[d]; int ny = j + dy[d]; if(inArea(nx, ny) && g[nx][ny] == 'X'){ st.x = min(i, nx), st.y = min(j, ny); st.lie = d < 2 ? 1 : 2; g[nx][ny] = g[i][j] = '.'; flag = true; } } if(!flag){ st.x = i, st.y = j, st.lie = 0; g[i][j] = '.'; } } } } } int bfs(){ memset(f, -1, sizeof f); queue<rec> q; f[st.x][st.y][st.lie] = 0; q.push(st); while(!q.empty()){ rec cur = q.front(); q.pop(); //cout << cur.x << " " << cur.y << endl; for(int i = 0; i < 4; i ++){ rec next; next.x = cur.x + nextx[cur.lie][i]; next.y = cur.y + nexty[cur.lie][i]; next.lie = nextlie[cur.lie][i]; if(valid(next) && f[next.x][next.y][next.lie] == -1){ f[next.x][next.y][next.lie] = f[cur.x][cur.y][cur.lie] + 1; if(next.x == ed.x && next.y == ed.y && next.lie == ed.lie) return f[next.x][next.y][next.lie]; q.push(next); } } } return -1; } int main(){ while(scanf("%d%d", &r, &c) != EOF && r && c){ for(int i = 0; i < r; i ++) scanf("%s", g[i]); init(); int ans = bfs(); if(ans == -1) printf("Impossible\n"); else printf("%d\n", ans); } return 0; }
POJ 3322 Bloxorz(算競進階習題)