poj 1475 || zoj 249 Pushing Boxes

http://poj.org/problem?id=1475
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=249
Pushing Boxes

Time Limit: 2000MS		Memory Limit: 131072K
Total Submissions: 4662		Accepted: 1608		Special Judge

Description

Imagine you are standing inside a two-dimensional maze composed of square cells which may or may not be filled with rock. You can move north, south, east or west one cell at a step. These moves are called walks.
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?

//解题思路：先判断盒子的四周是不是有空位，如果有，则判断人是否能到达盒子的那一边，能的话，把盒子推到空位处,然后继续
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#include<string>
#include<cmath>
using namespace std;
int bx,by,sx,sy,tx,ty;
int m,n,dir[][2]={-1,0,1,0,0,-1,0,1};//注意题目要求的是n、s、w、e的顺序，因为这个wa了一次
char op[]={'n','s','w','e'};
bool mark[25][25][4];//标记箱子四周的位置时候已被用过
int vis[25][25];//标记人走过的位置
char  map[25][25];
struct BB//盒子
{
int x,y,SX,SY;//SX,SY表示当前箱子固定了，人所在的位置
string ans;
}bnow,beed;
struct YY//人
{
int x,y;
string ans;
}ynow,yeed;
char up(char c)
{
return (c-'a'+'A');
}
//aa,bb 表示当前盒子的位置；ss，ee表示起点；
bool bfs2(int s,int e,int aa,int bb,int ss,int ee)//寻找当前人，是否能够到达盒子指定的位置；
{
queue<YY>yy;
if(s<0 || s>m || e<0 || e>n || map[s][e] == '#') return false;
ynow.x = ss; ynow.y = ee; ynow.ans="";
memset(vis,0,sizeof(vis));
vis[aa][bb] =1;//不能经过盒子
vis[ss][ee] = 1;//起点标记为
yy.push(ynow);
while(!yy.empty())
{
ynow = yy.front();
yy.pop();
if(ynow.x == s && ynow.y == e)
{
return true;
}
for(int i=0;i<4;i++)
{
yeed.x = ynow.x+dir[i][0];
yeed.y = ynow.y+dir[i][1];
if(yeed.x>0 && yeed.x<=m && yeed.y>0 && yeed.y<=n && !vis[yeed.x][yeed.y] && map[yeed.x][yeed.y]!='#')
{
yeed.ans = ynow.ans+op[i];//记录走过的路径
vis[yeed.x][yeed.y] = 1;
yy.push(yeed);
}
}
}
return false;
}

bool bfs1()
{
queue<BB>bb;
bnow.x = bx;bnow.y=by;bnow.ans="";
bnow.SX = sx;bnow.SY=sy;
bb.push(bnow);
while(!bb.empty())
{

bnow=bb.front();
bb.pop();
if(bnow.x == tx && bnow.y==ty)
{
return true;
}
for(int i=0;i<4;i++) //盒子周围的四个方向；
{
beed.x = bnow.x+dir[i][0];
beed.y = bnow.y+dir[i][1];
if(beed.x>0 && beed.x<=m && beed.y>0 && beed.y<=n && !mark[beed.x][beed.y][i] && map[beed.x][beed.y]!='#')
{
if(bfs2(beed.x-2*dir[i][0],beed.y-2*dir[i][1],bnow.x,bnow.y,bnow.SX,bnow.SY))//如果能推到yeed，则需要判断人是否能到达，它的上一个点；
{
beed.SX = bnow.x;//推完箱子后，人的位置在箱子上
beed.SY = bnow.y;
beed.ans=bnow.ans+ynow.ans+up(op[i]);//当前的加上推箱子的加上目前挨着推的；
mark[beed.x][beed.y][i] = true;
bb.push(beed);
}
}
}
}
return false;
}

int main()
{
int T,k=1;
while(scanf("%d %d",&m,&n) && m+n)
{
memset(mark,false,sizeof(mark));
for(int i=1;i<=m;i++)
for(int j=1;j<=n;j++)
{
scanf(" %c",&map[i][j]);
if(map[i][j] == 'S')
{
sx=i;sy =j;
}
if(map[i][j] == 'T')
{
tx = i;ty = j;
}
if(map[i][j] == 'B')
{
bx = i;by = j;
}
}
printf("Maze #%d\n",k++);
if(bfs1())
printf("%s\n\n",bnow.ans.c_str());//少个换行wa了一次
else
printf("Impossible.\n\n");
}
return 0;
}

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