深度优先搜索【POJ 3009】

Curling 2.0

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 18356 Accepted: 7548

Description

On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from ours. The game is played on an ice game board on which a square mesh is marked. They use only a single stone. The purpose of the game is to lead the stone from the start to the goal with the minimum number of moves.

Fig. 1 shows an example of a game board. Some squares may be occupied with blocks. There are two special squares namely the start and the goal, which are not occupied with blocks. (These two squares are distinct.) Once the stone begins to move, it will proceed until it hits a block. In order to bring the stone to the goal, you may have to stop the stone by hitting it against a block, and throw again.

Fig. 1: Example of board (S: start, G: goal)

The movement of the stone obeys the following rules:

· At the beginning, the stone stands still at the start square.

· The movements of the stone are restricted to x and y directions. Diagonal moves are prohibited.

· When the stone stands still, you can make it moving by throwing it. You may throw it to any direction unless it is blocked immediately(Fig. 2(a)).

· Once thrown, the stone keeps moving to the same direction until one of the following occurs:

o The stone hits a block (Fig. 2(b), (c)).

§ The stone stops at the square next to the block it hit.

§ The block disappears.

o The stone gets out of the board.

§ The game ends in failure.

o The stone reaches the goal square.

§ The stone stops there and the game ends in success.

· You cannot throw the stone more than 10 times in a game. If the stone does not reach the goal in 10 moves, the game ends in failure.

Fig. 2: Stone movements

Under the rules, we would like to know whether the stone at the start can reach the goal and, if yes, the minimum number of moves required.

With the initial configuration shown in Fig. 1, 4 moves are required to bring the stone from the start to the goal. The route is shown in Fig. 3(a). Notice when the stone reaches the goal, the board configuration has changed as in Fig. 3(b).

Fig. 3: The solution for Fig. D-1 and the final board configuration

Input

The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. The number of datasets never exceeds 100.

Each dataset is formatted as follows.

the width(=w) and the height(=h) of the board 
First row of the board 
... 
h-th row of the board

The width and the height of the board satisfy: 2 <= w <= 20, 1 <= h <= 20.

Each line consists of w decimal numbers delimited by a space. The number describes the status of the corresponding square.

0

vacant square

1

block

2

start position

3

goal position

The dataset for Fig. D-1 is as follows:

6 6 
1 0 0 2 1 0 
1 1 0 0 0 0 
0 0 0 0 0 3 
0 0 0 0 0 0 
1 0 0 0 0 1 
0 1 1 1 1 1

Output

For each dataset, print a line having a decimal integer indicating theminimum number of moves along a route from the start to the goal. If there are no such routes, print -1 instead. Each line should not have any character other than this number.

Sample Input

2 1

3 2

6 6

1 0 0 2 1 0

1 1 0 0 0 0

0 0 0 0 0 3

0 0 0 0 0 0

1 0 0 0 0 1

0 1 1 1 1 1

6 1

1 1 2 1 1 3

6 1

1 0 2 1 1 3

12 1

2 0 1 1 1 1 1 1 1 1 1 3

13 1

2 0 1 1 1 1 1 1 1 1 1 1 3

0 0

Sample Output

1

4

-1

4

10

-1

 思路：

(1)冰壶在没有遇到阻碍时，要求一直往前走，而深度搜索是每走一步可能会换方向，于是需要while循环，保持       方向不变也简单(nx += dx[i],ny += dy[i];)

(2)利用剪枝思想(在一次游戏中，扔冰壶不能超过10次，加快深搜的速度

(3)冰壶在滑动的过程中，遇到阻碍会停止并撞碎障碍物(A[nx][ny] = 0;),后退一步，可以选择之前没有障碍的路或      者继续走之前有障碍的路(障碍物以撞碎)

(4)回溯，假设在位置a，在其一方向前进一步撞碎障碍物后后退一步，仍有4种选择，而当这4方向遍历完，            应恢复之前被撞碎的障碍物，以保持现场

#include<iostream>#include<limits.h>using namespace std;const int MAX_W = 20,MAX_H = 20;int dx[4] = {1,0,-1,0},dy[4] = {0,1,0,-1};int A[MAX_H][MAX_W];int w,h;const int INF = 11;int sx,sy,minStep;int ok(int x,int y)//判断是否越界{    if(x >= 0 && x < h && y >= 0 && y < w)        return 1;    return 0;}void dfs(int step,int x,int y){    //cout << '[' << x << ',' << y << ']' << endl;//当前所在位置    if(step > 10)//剪枝(在一次游戏中，扔冰壶不能超过10次)        return;    for(int i = 0; i < 4; i++)    {        int nx = x + dx[i],ny = y + dy[i];        if(!ok(nx,ny) || A[nx][ny] == 1)//越界 || 眼前有障碍物在走不通            continue;        //cout << '{' << nx << ',' << ny << '}' << ' '<< ok(nx,ny) << endl;//扔到的第一块空地        while(!(A[nx][ny] == 1 || A[nx][ny] == 3))//如果没有遇到障碍物或没有到达终点，则沿着这方向一直往前滑        {            nx += dx[i],ny += dy[i];            //cout << '<' << nx << ',' << ny << '>' << ' ';            if(!ok(nx,ny))//越界            {                //cout << ok(nx,ny) << ' ';                break;            }        }        //结束while循环：(1)遇到障碍物 (2)到达终点 (3)越界        //cout << endl << '{' << nx << ',' << ny << '}' << endl;        if(!ok(nx,ny))//越界，直接退出本次循环，再换方向            continue;        if(A[nx][ny] == 3)//到达终点，计算步数        {            minStep = (minStep > (step+1)?(step+1):minStep);            //cout << "minStep:" << minStep << endl;        }        else if(A[nx][ny] == 1)//遇到障碍物，撞碎障碍物        {            A[nx][ny] = 0;            dfs(step+1,nx-dx[i],ny-dy[i]);            A[nx][ny] = 1; //回溯(先换方向，再恢复现场)            //cout << "A[" << nx << "][" << ny << "]" << endl;        }    }}int main(){    while(1)    {        cin >> w >> h;        if(!h && !w)            break;        for(int i = 0; i < h; i++)        {            for(int j = 0; j < w; j++)            {                cin >> A[i][j];                if(A[i][j] == 2)                {                    sx = i,sy = j;                    A[i][j] = 0;//起点是空闲场地，可以通过                }            }        }        minStep = INF;        dfs(0,sx,sy);        if(minStep <= 10)            cout << minStep << endl;        else            cout << -1 << endl;    }    return 0;}