POJ 3009 DFS+剪枝

POJ3009 DFS+剪枝

原题：

Curling 2.0

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 16280 Accepted: 6725

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:

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

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

The block disappears.

The stone gets out of the board.

The game ends in failure.

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 the minimum 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

题意：一个冰壶 求从起点“2”到终点“3”最少的步数。

“0”为可以移动的地方，“1”是障碍物，冰壶只有遇到了障碍物才能改变方向（冰壶停在障碍物的后面，障碍物变成可以移动的地方）&冰壶不可以出界&冰壶不可以刚出手就撞墙

跟真正的冰壶还是有点相似的。。。

---

没有选择每一步搜一次，觉得比较麻烦。还要把它分为“动过了”和“没有动过”两种状态。。

直接搜直线

给出代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[25][25],m,n,sx,sy,flag;
int xx[]={1,-1,0,0},yy[]={0,0,1,-1};
void dfs(int x,int y,int t)
{
    if(t>10||t>flag) return;//剪枝
    for(int i=0;i<=3;i++)
    {
        bool ok=false;
        int dx=x,dy=y;
        if(a[dx+xx[i]][dy+yy[i]]==1) continue;//不能一出手就撞墙啊
        while(1)//搜直线
        {
            if(a[dx+xx[i]][dy+yy[i]]==0||a[dx+xx[i]][dy+yy[i]]==2)
            {
                dx+=xx[i];dy+=yy[i];
            }
            else if(a[dx+xx[i]][dy+yy[i]]==1)
            {
                ok=true;
                break;
            }
            else if(a[dx+xx[i]][dy+yy[i]]==3)
            {
                if(flag>t) flag=t;//寻找最优解
                return;
            }
            else if(a[dx+xx[i]][dy+yy[i]]==-1) break;//越界
        }
        if(ok)
        {
            a[dx+xx[i]][dy+yy[i]]=0;
                dfs(dx,dy,t+1);//搜撞之前的
            a[dx+xx[i]][dy+yy[i]]=1;
        }
    }
}
int main()
{
    while(scanf("%d%d",&m,&n)&&n)//坑人的输入
    {
        flag=0x3f3f3f3f;
        memset(a,-1,sizeof(a));//懒得判越界了，直接赋值-1好了
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                scanf("%d",&a[i][j]);
                if(a[i][j]==2)
                {
                    sx=i;sy=j;
                }
            }
        }
        dfs(sx,sy,1);
        flag<=10?printf("%d\n",flag):printf("-1\n");
    }
}