POJ - 1475 Pushing Boxes

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题意：推箱子的题目，求最短的路径，如果推的最短一样的话，就算上走的最短

思路：首先我们记录状态：箱子的位置和人的位置，我们先BFS箱子的最短，然后我们可以通过推的方向得到人的位置，然后再BFS人是否能到这里的最短路

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int MAXN = 50;struct status{int person_x,person_y;int box_x,box_y;string ans;}p,q;struct node{int x,y;string ans;}P,Q;int dx[]={1, -1, 0, 0};int dy[]={0, 0, 1, -1};char Pm[]={'s', 'n', 'e', 'w'};char Pb[]={'S', 'N', 'E', 'W'};int visP[MAXN][MAXN],visB[MAXN][MAXN];char G[MAXN][MAXN];int sx,sy,ex,ey,bx,by;int r,c;int check(int x, int y){if (x >= 0 && x< r && y >= 0 && y < c)return true;return false;}int bfs2(int start_x, int start_y, int end_x, int end_y){memset(visP, 0, sizeof(visP));P.x = start_x;P.y = start_y;P.ans = "";visP[P.x][P.y] = 1;visP[p.box_x][p.box_y] = 1;queue<node> qq;qq.push(P);while (!qq.empty()){P = qq.front();qq.pop();if (P.x == end_x && P.y == end_y)return true;for (int i = 0; i < 4; i++){int nx = P.x + dx[i];int ny = P.y + dy[i];if (check(nx, ny) && G[nx][ny] != '#' && !visP[nx][ny]){visP[nx][ny] = 1;Q.ans = P.ans + Pm[i];Q.x = nx;Q.y = ny;qq.push(Q);}}}return false;}int bfs(){memset(visB, 0, sizeof(visB));p.person_x = sx;p.person_y = sy;p.box_x = bx;p.box_y = by;p.ans = "";visB[bx][by] = 1;queue<status> st;st.push(p);while (!st.empty()){p = st.front();st.pop();for (int i = 0; i < 4; i++){int nx = p.box_x + dx[i];int ny = p.box_y + dy[i];int tx = p.box_x - dx[i];int ty = p.box_y - dy[i];if (check(tx, ty) && G[tx][ty] != '#' && check(nx, ny) &&G[nx][ny] != '#' && !visB[nx][ny]){if (bfs2(p.person_x, p.person_y, tx, ty)){visB[nx][ny] = 1;q.person_x = p.box_x;q.person_y = p.box_y;q.box_x = nx;q.box_y = ny;q.ans = p.ans+P.ans+Pb[i];if (nx == ex && ny == ey)return true;st.push(q);}}}}return false;}int main(){int cas = 1;while (scanf("%d%d", &r, &c) != EOF && r+c){for (int i = 0; i < r; i++){scanf("%s", G[i]);for (int j = 0; j < c; j++){if (G[i][j] == 'S'){sx = i;sy = j;}if (G[i][j] == 'T'){ex = i;ey = j;}if (G[i][j] == 'B'){bx = i;by = j;}}}printf("Maze #%d\n", cas++);if (bfs())printf("%s\n", q.ans.c_str());else printf("Impossible.\n");printf("\n");}return 0;}

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