joj1113

 1113: The Game

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	380	97	Standard

One morning, you wake up and think: ``I am such a good programmer. Why not make some money?'' So you decide to write a computer game.

The game takes place on a rectangular board consisting of w * h squares. Each square might or might not contain a game piece, as shown in the picture.

One important aspect of the game is whether two game pieces can be connected by a path which satisfies the two following properties:

It consists of straight segments, each one being either horizontal or vertical. 

It does not cross any other game pieces. 

(It is allowed that the path leaves the board temporarily.)

Here is an example:

The game pieces at (1,3) and at (4, 4) can be connected. The game pieces at (2, 3) and (3, 4) cannot be connected; each path would cross at least one other game piece.

The part of the game you have to write now is the one testing whether two game pieces can be connected according to the rules above.

Input

The input contains descriptions of several different game situations. The first line of each description contains two integers w and h (1 <= w,h <= 75), the width and the height of the board. The next h lines describe the contents of the board; each of these lines contains exactly w characters: a ``X'' if there is a game piece at this location, and a space if there is no game piece.

Each description is followed by several lines containing four integers x1, y1, x2, y2 each satisfying 1 <= x1,x2 <= w, 1 <= y1,y2 <= h. These are the coordinates of two game pieces. (The upper left corner has the coordinates (1, 1).) These two game pieces will always be different. The list of pairs of game pieces for a board will be terminated by a line containing ``0 0 0 0".

The entire input is terminated by a test case starting with w=h=0. This test case should not be procesed.

Output

For each board, output the line ``Board #n:'', where n is the number of the board. Then, output one line for each pair of game pieces associated with the board description. Each of these lines has to start with ``Pair m: '', where m is the number of the pair (starting the count with 1 for each board). Follow this by ``ksegments.'', where k is the minimum number of segments for a path connecting the two game pieces, or ``impossible.'', if it is not possible to connect the two game pieces as described above.

Output a blank line after each board.

Sample Input

5 4
XXXXX
X   X
XXX X
 XXX
2 3 5 3
1 3 4 4
2 3 3 4
0 0 0 0
0 0

Sample Output

Board #1:
Pair 1: 4 segments.
Pair 2: 3 segments.
Pair 3: impossible.

This problem is used for contest: 149 

Submit / Problem List / Status / Discuss

这道题跟2558差不多都是一种特殊的队列应用。。
我感觉这道题的难点在于如何输出。我是首先在主函数中
首先对一些特殊情况进行输出从而减轻bfs的压力。
然后在bfs里面 有两个输出函数分别应对不同的情况。
第一个函数是为了能够在一个直通路上对通路两侧的点
进行分析判断有没有目标点
第二个函数目的在于对通路的尽头那个点进行分析判断是不是目标点。。
（我感觉我没有说清楚如果那个地方不明白可以给我留言。）




#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
using namespace std;
char map[100][100];
int h,w;
int visited[100][100];
bool success;
class Node
{
public:
int x,y;
int step;
};
bool ok(int x,int y,Node node)
{
//up
if(x-1==node.x&&y==node.y&&map[node.x][node.y]=='X')return true;
//down
if(x+1==node.x&&y==node.y&&map[node.x][node.y]=='X')return true;
//left
if(x==node.x&&y-1==node.y&&map[node.x][node.y]=='X')return true;
//right
if(x==node.x&&y+1==node.y&&map[node.x][node.y]=='X')return true;
return false;
}
int move[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
void bfs(Node node1,Node node2)
{
node1.step=0;
queue<Node>q;
q.push(node1);
visited[node1.x][node1.y]=1;
while(!q.empty())
{
Node temp;
temp=q.front();
q.pop();
if(ok(temp.x,temp.y,node2))
{
printf("%d segments.\n",temp.step+1);
success=false;
return ;
}
node1=temp;
node1.step++;
int x,y;
for(int i=0;i<4;i++)
{
temp=node1;
while(1)
{
x=temp.x+move[i][0];
y=temp.y+move[i][1];
if(x>=0&&x<=h+1&&y>=0&&y<=w+1&&!visited[x][y]&&map[x][y]!='X')
{
visited[x][y]=1;
temp.x=x;
temp.y=y;
temp.step=node1.step;
q.push(temp);
if(x+move[i][0]==node2.x&&y+move[i][1]==node2.y
   &&map[node2.x][node2.y]=='X')
{
printf("%d segments.\n",temp.step);
success=false;
return ;
}//对那一些
}
else break;
};
}
}
}
int main()
{
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
int n=1;
Node node1,node2;
while(scanf("%d%d",&w,&h),w&&h)
{
memset(map,' ',sizeof(map));
getchar();
for(int i=1;i<=h;i++)
{
for(int j=1;j<=w;j++)
{
map[i][j]=getchar();
}
getchar();
}
printf("Board #%d:\n",n++);
int m=1;
while(1)
{
memset(visited,0,sizeof(visited));
success=true;
scanf("%d%d%d%d",&node1.y,&node1.x,&node2.y,&node2.x);
if(node1.x==0&&node1.y==0&&node2.x==0&&node2.y==0)
break;
printf("Pair %d: ",m++);
if(node1.x==node2.x&&node1.y==node2.y&&map[node1.x][node1.y]=='X')
{printf("0 segments.\n");success=false;}
else if(map[node1.x][node1.y]!='X'||map[node2.x][node2.y]!='X')
{
printf("impossible.\n");
success=false;
}//首先我需要对一些简单情况进行分析从而能够在bfs减少负担
else bfs(node1,node2);
if(success)printf("impossible.\n");
}
printf("\n");
}
return 0;
}