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poj 1101 The Game(bfs)

編輯：C++入門知識

The Game Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 8640 Accepted: 2632

Description

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 accZ喎?http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">vcmRpbmcgdG8gdGhlIHJ1bGVzIGFib3ZlLgo8cCBjbGFzcz0="pst">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.

Source

Mid-Central European Regional Contest 1999

題意：

連連看的規則。問兩個卡之間的最少拐彎數。

思路：

很容易想到bfs但是寫到一半時卡了下。因為想到一格一格的走的話不能保證拐彎數是遞增的。還想寫個優先隊列。但是先明白了其實很簡單。這裡的一步已不再是一個。而是一條線。所以每次處理一個方向的一條線就行了。

詳細見代碼：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double PI=acos(-1.0);
const int maxn=100010;
typedef __int64 ll;
int dx[4]={-1,0,1,0};
int dy[4]={0,1,0,-1};
int w,h,head,tail,sx,sy,ex,ey,ans;
char maze[100][100];
int vis[100][100];
struct yb
{
    int x,y;
    int dre,sp;
} q[10000];
void bfs()
{
    int i,x,y,nx,ny,dr;
    memset(vis,0,sizeof vis);
    vis[sx][sy]=1;
    head=tail=0;
    q[tail].x=sx;
    q[tail].y=sy;
    q[tail].dre=4;
    q[tail++].sp=0;
    while(head=0&&nx<=h+1&&ny>=0&&ny<=w+1&&!vis[nx][ny]&&maze[nx][ny]!='X')//一次處理一條線
            {
                vis[nx][ny]=1;
                q[tail].x=nx;
                q[tail].y=ny;
                q[tail].sp=q[head].sp+1;
                q[tail].dre=i;
                tail++;
                nx+=dx[i];
                ny+=dy[i];
            }
            if(nx==ex&&ny==ey)
            {
                ans=q[head].sp+1;
                return;
            }
        }
        head++;
    }
}
int main()
{
    int i,bd=1,pa;
    while(scanf("%d%d",&w,&h),w||h)
    {
        pa=1;
        getchar();
        memset(maze,0,sizeof maze);
        for(i=1;i<=h;i++)
            gets(maze[i]+1);
        printf("Board #%d:\n",bd++);
        while(scanf("%d%d%d%d",&sy,&sx,&ey,&ex),sx||sy||ex||ey)
        {
            ans=-1;
            bfs();
            if(ans!=-1)
                printf("Pair %d: %d segments.\n",pa++,ans);
            else
                printf("Pair %d: impossible.\n",pa++);
        }
        printf("\n");
    }
    return 0;
}