[POJ 1475] Pushing Boxes【A*】
Problem:
Time Limit: 2000MS | Memory Limit: 131072K | Special Judge |
Description
One of the empty cells contains a box which can be moved to an adjacent free cell by standing next to the box and then moving in the direction of the box. Such a move is called a push. The box cannot be moved in any other way than by pushing, which means that if you push it into a corner you can never get it out of the corner again.One of the empty cells is marked as the target cell. Your job is to bring the box to the target cell by a sequence of walks and pushes. As the box is very heavy, you would like to minimize the number of pushes. Can you write a program that will work out the best such sequence?

Input
The input contains the descriptions of several mazes. Each maze description starts with a line containing two integers r and c (both <= 20) representing the number of rows and columns of the maze.
Following this are r lines each containing c characters. Each character describes one cell of the maze. A cell full of rock is indicated by a `#' and an empty cell is represented by a `.'. Your starting position is symbolized by `S', the starting position of the box by `B' and the target cell by `T'.
Input is terminated by two zeroes for r and c.
Output
For each maze in the input, first print the number of the maze, as shown in the sample output. Then, if it is impossible to bring the box to the target cell, print ``Impossible.''.
Otherwise, output a sequence that minimizes the number of pushes. If there is more than one such sequence, choose the one that minimizes the number of total moves (walks and pushes). If there is still more than one such sequence, any one is acceptable.
Print the sequence as a string of the characters N, S, E, W, n, s, e and w where uppercase letters stand for pushes, lowercase letters stand for walks and the different letters stand for the directions north, south, east and west.
Output a single blank line after each test case.
Sample Input
1 7 SB....T 1 7 SB..#.T 7 11 ########### #T##......# #.#.#..#### #....B....# #.######..# #.....S...# ########### 8 4 .... .##. .#.. .#.. .#.B .##S .... ###T 0 0
Sample Output
Maze #1 EEEEE Maze #2 Impossible. Maze #3 eennwwWWWWeeeeeesswwwwwwwnNN Maze #4 swwwnnnnnneeesssSSS
Source
Solution:
一道非常不错的 A* 大水题。
采用 Manhattan 距离作为估价函数,将 push 操作和 walk 操作分开评估:
- push: h() = Manhattan(box, destination)。
- walk: h() = Manhattan(man, box) - 1,减 1 的原因是 man 只要走到 box 四周的某一格即可。
此题的“关闭列表”改为刚得到状态时直接记录最小代价,而不是当状态从开启列表中移除后记录,可以减少入队次数从而提升时空效率。
路径记录数组一定要开得足够大。
Code: O(玄学) [2324K, 79MS]
#include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<queue> using namespace std; const int dx[4] = {0, 1, 0,-1}; const int dy[4] = {1, 0,-1, 0}; const char DIR[4] = {'E', 'S', 'W', 'N'}; int cas = 0, R, C; char mp[22][22]; struct Point{ int x, y; inline bool operator == (const Point &p2) const {return x == p2.x && y == p2.y;} inline bool isLegal() {return x > 0 && x <= R && y > 0 && y <= C && mp[x][y] != '#';} } T; struct State{ Point man, box; int pf, pg, ph, wf, wg, wh; int id; inline bool operator < (const State &st2) const { if(pf != st2.pf) return pf > st2.pf; return wf > st2.wf; } } S; priority_queue<State> openList; int closeList[22][22][22][22][2]; inline void clearPQ(priority_queue<State> &pq) {priority_queue<State> tmp; swap(tmp, pq);} inline int Manhattan(const Point &p1, const Point &p2) {return abs(p1.x - p2.x) + abs(p1.y - p2.y);} inline bool isBetter(const State &st){ int *buf = closeList[st.man.x][st.man.y][st.box.x][st.box.y]; if(buf[0] != st.pf) return st.pf < buf[0]; return st.wf < buf[1]; } inline void updateClose(const State &st){ int *buf = closeList[st.man.x][st.man.y][st.box.x][st.box.y]; buf[0] = st.pf, buf[1] = st.wf; } int fromId[100001], topfrom; char fromDir[100001]; inline int Astar(){ S.pg = 0, S.ph = Manhattan(S.box, T), S.pf = S.pg + S.ph; S.wg = 0, S.wh = Manhattan(S.man, S.box) - 1, S.wf = S.wg + S.wh; S.id = 0; openList.push(S); while(!openList.empty()){ State u = openList.top(), v; openList.pop(); for(register int i = 0; i < 4; i++){ v = u, v.man.x += dx[i], v.man.y += dy[i]; if(!v.man.isLegal()) continue; if(v.man == v.box){ // Push v.box.x += dx[i], v.box.y += dy[i]; if(!v.box.isLegal()) continue; v.pg++, v.ph = Manhattan(v.box, T), v.pf = v.pg + v.ph; if(!isBetter(v)) continue; fromId[v.id = ++topfrom] = u.id, fromDir[v.id] = DIR[i]; openList.push(v); updateClose(v); if(v.box == T) return v.id; } else{ // Walk v.wg++, v.wh = Manhattan(v.man, v.box) - 1, v.wf = v.wg + v.wh; if(!isBetter(v)) continue; fromId[v.id = ++topfrom] = u.id, fromDir[v.id] = DIR[i] - 'A' + 'a'; openList.push(v); updateClose(v); } } } return -1; } inline void traceBack(int cur){ if(fromId[cur]) traceBack(fromId[cur]); putchar(fromDir[cur]); } int main(){ while(scanf("%d%d", &R, &C) != EOF && R && C){ for(register int i = 1; i <= R; i++){ scanf("%s", mp[i] + 1); for(register int j = 1; j <= C; j++){ if(mp[i][j] == 'T') T.x = i, T.y = j; if(mp[i][j] == 'S') S.man.x = i, S.man.y = j; if(mp[i][j] == 'B') S.box.x = i, S.box.y = j; } } clearPQ(openList), memset(closeList, 0x3f, sizeof(closeList)); topfrom = 0; int endId = Astar(); printf("Maze #%d\n", ++cas); if(endId == -1) puts("Impossible.\n"); else traceBack(endId), puts("\n"); } return 0; }
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