POJ-1475（A*算法）

题目：http://poj.org/problem?id=1475

一眼看上去就能确定是BFS，但这题和普通寻找终点的BFS不一样，除了自身的位置，状态上还要体现箱子的位置。由于题目要求最少push最少walk，因此可以A*。

一个重要的问题是，搜到了之后如何生成路径，可以这么考虑：A*搜索使得我们到达终点完成任务时，走的路径时最优的，由于状态间是一步转移的，因此上一个状态形成的路径也肯定是最优的，这样一直找到最优的起点push = 0, walk = 0，由于搜索时已经保存了各个最优状态（用于剪枝），因此可以递归的打印路径。

一开始犯了个错误，在生成路径时没有完整考虑推操作和走操作需要的条件WA了一次，推操作箱子的上一个位置及人现在的位置，而走操作的上一位置不能和箱子现在的位置一样，看了Discuss中的数据发现问题所在，改了之后即A了。

#include <cstdio>#include <cctype>#include <cstring>#include <queue>using namespace std;const int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};const char reverseDir[4] = {'S', 'N', 'E', 'W'};int R, C;char map[20][21];int best[20][20][20][20][2];struct State{int boxRow, boxCol, manRow, manCol, push, walk;bool operator < (const State& other)const{if(push != other.push) return push > other.push;return walk > other.walk;}void show()const{printf("box@(%d,%d)  man@(%d,%d)  push:%d  walk:%d\n", boxRow, boxCol, manRow, manCol, push, walk);}};priority_queue<State> Q;inline bool isUreachable(int r, int c){return r < 0 || r >= R || c < 0 || c >= C || map[r][c] == '#';}inline bool isBetter(const State& state){int* a = best[state.boxRow][state.boxCol][state.manRow][state.manCol];if(a[0] != state.push) return a[0] > state.push;return a[1] > state.walk;}inline void setBest(const State& state){int* a = best[state.boxRow][state.boxCol][state.manRow][state.manCol];a[0] = state.push;a[1] = state.walk;}void printPath(int br, int bc, int mr, int mc, int p, int w){if(p == 0 && w == 0) return;for(int i = 0; i < 4; ++i){int pmr = mr + dir[i][0], pmc = mc + dir[i][1];if(isUreachable(pmr, pmc)) continue;//if this move is walk, previously man can not be at current box's placeif(!(pmr == br && pmc == bc) && best[br][bc][pmr][pmc][0] == p && best[br][bc][pmr][pmc][1] == w-1){printPath(br, bc, pmr, pmc, p, w-1);putchar(tolower(reverseDir[i]));break;}//if this move is push, previously box is at current man's placeint pbr = br + dir[i][0], pbc = bc + dir[i][1];if(isUreachable(pbr, pbc)) continue;if(pbr == mr && pbc == mc && best[pbr][pbc][pmr][pmc][0] == p-1 && best[pbr][pbc][pmr][pmc][1] == w){printPath(pbr, pbc, pmr, pmc, p-1, w);putchar(reverseDir[i]);break;}}}void solve(){//initializewhile(!Q.empty()) Q.pop();memset(best, 0x6F, sizeof(best));//find startState now, nex;for(int i = 0; i < R; ++i){char* p = strchr(map[i], 'B');if(p){now.boxRow = i;now.boxCol = p - map[i];}p = strchr(map[i], 'S');if(p){now.manRow = i;now.manCol = p - map[i];}}now.push = now.walk = 0;//AStarQ.push(now);setBest(now);while(!Q.empty()){now = Q.top(); Q.pop();//now.show();if(map[now.boxRow][now.boxCol] == 'T'){printPath(now.boxRow, now.boxCol, now.manRow, now.manCol, now.push, now.walk);puts("");return;}for(int i = 0; i < 4; ++i){nex = now;nex.manRow += dir[i][0];nex.manCol += dir[i][1];if(isUreachable(nex.manRow, nex.manCol)) continue;if(nex.manRow == now.boxRow && nex.manCol == now.boxCol){//this move is a push//check if box can move or visit this state beforenex.boxRow += dir[i][0];nex.boxCol += dir[i][1];if(isUreachable(nex.boxRow, nex.boxCol)) continue;++nex.push;if(!isBetter(nex)) continue;Q.push(nex);setBest(nex);}else{//this move is a walk++nex.walk;if(!isBetter(nex)) continue;Q.push(nex);setBest(nex);}}}puts("Impossible.");}int main(){int kase = 0, i;while(scanf("%d%d", &R, &C), R){while(getchar() != '\n') ;for(i = 0; i < R; ++i) gets(map[i]);printf("Maze #%d\n", ++kase);solve();puts("");}return 0;}