【BFS】The Morning after Halloween

[POJ3523]The Morning after Halloween

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 2395		Accepted: 543

Description

You are working for an amusement park as an operator of an obakeyashiki, or a haunted house, in which guests walk through narrow and dark corridors. The house is proud of their lively ghosts, which are actually robots remotely controlled by the operator, hiding here and there in the corridors. One morning, you found that the ghosts are not in the positions where they are supposed to be. Ah, yesterday was Halloween. Believe or not, paranormal spirits have moved them around the corridors in the night. You have to move them into their right positions before guests come. Your manager is eager to know how long it takes to restore the ghosts.

In this problem, you are asked to write a program that, given a floor map of a house, finds the smallest number of steps to move all ghosts to the positions where they are supposed to be.

A floor consists of a matrix of square cells. A cell is either a wall cell where ghosts cannot move into or a corridor cell where they can.

At each step, you can move any number of ghosts simultaneously. Every ghost can either stay in the current cell, or move to one of the corridor cells in its 4-neighborhood (i.e. immediately left, right, up or down), if the ghosts satisfy the following conditions:

No more than one ghost occupies one position at the end of the step.
No pair of ghosts exchange their positions one another in the step.

For example, suppose ghosts are located as shown in the following (partial) map, where a sharp sign (‘#’) represents a wall cell and ‘a’, ‘b’, and ‘c’ ghosts.

####
 ab#
#c##
####

The following four maps show the only possible positions of the ghosts after one step.

####
 ab#
#c##
####

####
a b#
#c##
####

####
acb#
# ##
####

####
ab #
#c##
####

Input

The input consists of at most 10 datasets, each of which represents a floor map of a house. The format of a dataset is as follows.

w	h	n
c₁₁	c₁₂	⋯	c_1w
c₂₁	c₂₂	⋯	c_2w
⋮	⋮	⋱	⋮
c_h1	c_h2	⋯	c_hw

w, h and n in the first line are integers, separated by a space. w and h are the floor width and height of the house, respectively. n is the number of ghosts. They satisfy the following constraints.

4 ≤ w ≤ 16, 4 ≤ h ≤ 16, 1 ≤ n ≤ 3

Subsequent h lines of w characters are the floor map. Each of c_ij is either:

a ‘#’ representing a wall cell,
a lowercase letter representing a corridor cell which is the initial position of a ghost,
an uppercase letter representing a corridor cell which is the position where the ghost corresponding to its lowercase letter is supposed to be, or
a space representing a corridor cell that is none of the above.

In each map, each of the first n letters from a and the first n letters from A appears once and only once. Outermost cells of a map are walls; i.e. all characters of the first and last lines are sharps; and the first and last characters on each line are also sharps. All corridor cells in a map are connected; i.e. given a corridor cell, you can reach any other corridor cell by following corridor cells in the 4-neighborhoods. Similarly, all wall cells are connected. Any 2 × 2 area on any map has at least one sharp. You can assume that every map has a sequence of moves of ghosts that restores all ghosts to the positions where they are supposed to be.

The last dataset is followed by a line containing three zeros separated by a space.

Output

For each dataset in the input, one line containing the smallest number of steps to restore ghosts into the positions where they are supposed to be should be output. An output line should not contain extra characters such as spaces.

Sample Input

5 5 2

#####

#A#B#

#   #

#b#a#

#####

16 4 3

################

## ########## ##

#    ABCcba    #

################

16 16 3

################

### ##    #   ##

##  #  ##   # c#

#  ## ########b#

# ##  # #   #  #

#  # ##   # # ##

##  a#  # # #  #

### ## #### ## #

##   #   #  #  #

#  ##### # ## ##

####   #B# #   #

##  C#   #   ###

#  # # ####### #

# ######  A##  #

#        #    ##

################

0 0 0

Sample Output

Source

Japan 2007

题目大意：详见《算法入门经典》第7章7-9 万圣节的早晨(P205)

　　　　　就是有一个M*N的地图，有K个字母，小写字母要走到大写字母去，可以同时移动，但移动后不能重叠或交换，每2*2的矩形中保证至少会有2个是障碍，问最小步数。

试题分析：这道题目非常好，一道费劲的BFS，但是很值得写的。写死我了

　　　　　 vis[17][17][17][17][17][17]，普通BFS是最显然的方法，但会炸的很惨。。。

　　　　　那么有终止和起始，我们可以考虑用双向BFS。

　　　　　但这样写普通的双向BFS会炸空间，所以还是要优化。

　　　　　如何优化呢？yy了一个想法：将可以走的地方全标号(按顺序)

　　　　　那么我们只需要开vis[151][151][151]就足以标记了。

　　　　　然后有一点要注意一下，就是双向BFS第一个搜出来的不一定是最优解。

　　　　　要将同层的先搜完再确定答案。

代码：

#include<iostream>

#include<cstring>

#include<cstdio>

#include<vector>

#include<algorithm>

using namespace std;

inline int read(){

	int x=0,f=1;char c=getchar();

	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;

	for(;isdigit(c);c=getchar()) x=x*10+c-'0';

	return x*f;

}

const int INF = 9999999;

const int MAXN = 100000;

int N,M,K;

char MP[17][17];

int Ax,Ay; int Bx,By; int Cx,Cy;

int ax,ay; int bx,by; int cx,cy;

int vis[151][151][151];

int visB[151][151][151];

struct data{

	int ax,ay;

	int bx,by;

	int cx,cy;

	int stp;

}que[1000001];

struct data2{

	int Ax,Ay;

	int Bx,By;

	int Cx,Cy;

	int stp;

}que2[1000001];

int dis[6][2]={{0,1},{1,0},{0,-1},{-1,0},{0,0}};

int ls=1,rs=0;

int lb=1,rb=0;

int dit[21][21];

void in(int axx,int ayy,int bxx,int byy,int cxx,int cyy,int sp,bool fg){

	if(!fg){

		vis[dit[axx][ayy]][dit[bxx][byy]][dit[cxx][cyy]]=sp;

		que[++rs].ax=axx;

		que[rs].ay=ayy;

		que[rs].bx=bxx;

		que[rs].by=byy;

		que[rs].cx=cxx;

		que[rs].cy=cyy;

		que[rs].stp=sp;

	}

	else{

		visB[dit[axx][ayy]][dit[bxx][byy]][dit[cxx][cyy]]=sp;

		que2[++rb].Ax=axx;

		que2[rb].Ay=ayy;

		que2[rb].Bx=bxx;

		que2[rb].By=byy;

		que2[rb].Cx=cxx;

		que2[rb].Cy=cyy;

		que2[rb].stp=sp;

	}

}

int ans;

bool flag=false;

void BFS(){

	in(ax,ay,bx,by,cx,cy,0,0);

	in(Ax,Ay,Bx,By,Cx,Cy,0,1);

	int x1,y1,x2,y2,x3,y3,sp;

	int t=0;

	while(ls<=rs&&lb<=rb){

		if(flag) t++;

		if(t>1024) break;

		x1=que[ls].ax;

		y1=que[ls].ay;

		x2=que[ls].bx;

		y2=que[ls].by;

		x3=que[ls].cx;

		y3=que[ls].cy;

		sp=que[ls].stp;

		//cout<<"S:"<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<" "<<x3<<" "<<y3<<" "<<sp<<endl;

		for(int i=0;i<5;i++){

			int xx1=x1+dis[i][0];

			int yy1=y1+dis[i][1];

			if(xx1>N||yy1>M||yy1<1||xx1<1) continue;

			if(MP[xx1][yy1]=='#') continue;

			if(x2||y2)

				for(int j=0;j<5;j++){

					int xx2=x2+dis[j][0];

					int yy2=y2+dis[j][1];

					if(xx2==xx1&&yy1==yy2) continue;

					if(xx2>N||yy2>M||xx2<1||yy2<1) continue;

					if(xx2==x1&&yy2==y1&&xx1==x2&&yy1==y2) continue;

					if(MP[xx2][yy2]=='#') continue;

					if(x3||y3)

						for(int k=0;k<5;k++){

							if(i==4&&j==4&&k==4) continue;

							int xx3=x3+dis[k][0];

				    		int yy3=y3+dis[k][1];

				    		if(xx2==x3&&yy2==y3&&xx3==x2&&yy3==y2) continue;

				    		if(xx3==x1&&yy3==y1&&xx1==x3&&yy1==y3) continue;

				    		if((xx3==xx1&&yy3==yy1)||(xx3==xx2&&yy3==yy2)) continue;

				    		if(xx3>N||yy3>M||xx3<1||yy3<1||vis[dit[xx1][yy1]][dit[xx2][yy2]][dit[xx3][yy3]]!=-1) continue;

							if(MP[xx3][yy3]=='#') continue;

							if(visB[dit[xx1][yy1]][dit[xx2][yy2]][dit[xx3][yy3]]!=-1){

				    			ans=min(visB[dit[xx1][yy1]][dit[xx2][yy2]][dit[xx3][yy3]]+sp+1,ans);

				    			flag=true;// return ;

				    		}

				    		in(xx1,yy1,xx2,yy2,xx3,yy3,sp+1,0);

						}

					else{

						if(vis[dit[xx1][yy1]][dit[xx2][yy2]][dit[x3][y3]]!=-1) continue;

						if(i==4&&j==4) continue;

						//cout<<"tmp:"<<xx1<<" "<<yy1<<" "<<xx2<<" "<<yy2<<endl;

						if(visB[dit[xx1][yy1]][dit[xx2][yy2]][dit[x3][y3]]!=-1){

				    		ans=min(visB[dit[xx1][yy1]][dit[xx2][yy2]][dit[x3][y3]]+sp+1,ans);

				    		flag=true; //return ;

				    	}

				    	in(xx1,yy1,xx2,yy2,x3,y3,sp+1,0);

					}

				}

			else{

				if(vis[dit[xx1][yy1]][dit[x2][y2]][dit[x3][y3]]!=-1) continue;

				if(i==4) continue;

				if(visB[dit[xx1][yy1]][dit[x2][y2]][dit[x3][y3]]!=-1){

				    ans=min(visB[dit[xx1][yy1]][dit[x2][y2]][dit[x3][y3]]+sp+1,ans);

				    flag=true; //return ;

				}

				in(xx1,yy1,x2,y2,x3,y3,sp+1,0);

			}

		}

		x1=que2[lb].Ax;

		y1=que2[lb].Ay;

		x2=que2[lb].Bx;

		y2=que2[lb].By;

		x3=que2[lb].Cx;

		y3=que2[lb].Cy;

		sp=que2[lb].stp;

		//cout<<"B:"<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<" "<<x3<<" "<<y3<<" "<<sp<<endl;

		for(int i=0;i<5;i++){

			int xx1=x1+dis[i][0];

			int yy1=y1+dis[i][1];

			if(xx1>N||yy1>M||yy1<1||xx1<1) continue;

			if(MP[xx1][yy1]=='#') continue;

			if(x2||y2)

				for(int j=0;j<5;j++){

					int xx2=x2+dis[j][0];

					int yy2=y2+dis[j][1];

					if(xx2==x1&&yy2==y1&&xx1==x2&&yy1==y2) continue;

					if(xx2==xx1&&yy1==yy2) continue;

					if(xx2>N||yy2>M||xx2<1||yy2<1) continue;

					if(MP[xx2][yy2]=='#') continue;

					if(x3||y3)

						for(int k=0;k<5;k++){

							if(i==4&&j==4&&k==4) continue;

							int xx3=x3+dis[k][0];

				    		int yy3=y3+dis[k][1];

				    		if(xx2==x3&&yy2==y3&&xx3==x2&&yy3==y2) continue;

				    		if(xx3==x1&&yy3==y1&&xx1==x3&&yy1==y3) continue;

				    		if((xx3==xx1&&yy3==yy1)||(xx3==xx2&&yy3==yy2)) continue;

				    		if(xx3>N||yy3>M||xx3<1||yy3<1||visB[dit[xx1][yy1]][dit[xx2][yy2]][dit[xx3][yy3]]!=-1) continue;

				    		if(MP[xx3][yy3]=='#') continue;

							if(vis[dit[xx1][yy1]][dit[xx2][yy2]][dit[xx3][yy3]]!=-1){

				    			ans=min(vis[dit[xx1][yy1]][dit[xx2][yy2]][dit[xx3][yy3]]+sp+1,ans);

				    			flag=true; //return ;

				    		}

				    		in(xx1,yy1,xx2,yy2,xx3,yy3,sp+1,1);

						}

					else{

						if(visB[dit[xx1][yy1]][dit[xx2][yy2]][dit[x3][y3]]!=-1) continue;

						if(i==4&&j==4) continue;

						//cout<<"tmp:"<<xx1<<" "<<yy1<<" "<<xx2<<" "<<yy2<<" "<<sp+1<<endl;

						if(vis[dit[xx1][yy1]][dit[xx2][yy2]][dit[x3][y3]]!=-1){

				    		ans=min(vis[dit[xx1][yy1]][dit[xx2][yy2]][dit[x3][y3]]+sp+1,ans);

				    		flag=true; //return ;

				    	}

				    	in(xx1,yy1,xx2,yy2,x3,y3,sp+1,1);

					}

				}

			else{

				if(visB[dit[xx1][yy1]][dit[x2][y2]][dit[x3][y3]]!=-1) continue;

				if(i==4) continue;

				if(vis[dit[xx1][yy1]][dit[x2][y2]][dit[x3][y3]]!=-1){

				    ans=min(vis[dit[xx1][yy1]][dit[x2][y2]][dit[x3][y3]]+sp+1,ans);

				    flag=true; //return ;

				}

				in(xx1,yy1,x2,y2,x3,y3,sp+1,1);

			}

		}

		ls++,lb++;

	}

}

int main(){

	//freopen("a.txt","w",stdout);

	while(1){

   		M=read(),N=read(),K=read();

    	if(!N||!M) break;

    	Ax=Ay=Bx=By=Cx=Cy=ax=ay=bx=by=cx=cy=0;

    	ls=1,rs=0;

		lb=1,rb=0;

		for(int i1=0;i1<=150;i1++)

		    for(int i2=0;i2<=150;i2++)

		        for(int i3=0;i3<=150;i3++){

		        	vis[i1][i2][i3]=-1;

		        	visB[i1][i2][i3]=-1;

		        }

		int cnt=0;

		for(int i=1;i<=N;i++){

        	for(int j=1;j<=M;j++){

        		MP[i][j]=getchar();

        	    if(MP[i][j]=='A') Ax=i,Ay=j;

        	    if(MP[i][j]=='B') Bx=i,By=j;

        	    if(MP[i][j]=='C') Cx=i,Cy=j;

        	    if(MP[i][j]=='a') ax=i,ay=j;

        	    if(MP[i][j]=='b') bx=i,by=j;

        	    if(MP[i][j]=='c') cx=i,cy=j;

        	    if(MP[i][j]!='#'){

        	    	dit[i][j]=++cnt;

        	    }

			}

		    getchar();

		}

		ans=INF; flag=false;

		BFS();

		cout<<ans<<endl;

    }

}

巴特西

【BFS】The Morning after Halloween

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