ACM Curling 2.0(挑战程序设计竞赛)_curling 2.0 java-优快云博客

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在MM-21星球上，一种类似冰壶的运动正在流行，但规则有所不同。玩家需要将石头从起点推到终点，途中可能需要利用阻挡物进行转向，最少步数为胜利。游戏结束条件包括石头超过10次移动未到达终点或成功抵达终点。输入包含关卡地图，输出是最小移动次数，若无法到达则输出-1。

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Curling 2.0

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3009
64-bit integer IO format: %lld Java class name: Main

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Type:

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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

Source

Japan 2006 Domestic

题目大意：一个冰壶游戏．开始时冰壶在Ｓ处，你可以向上下左右四个方向推冰壶，前提是那个方向没有紧挨这一个物块．

冰壶碰到物块会停在那个物块前，并且碰到的那个物块会小时．

注意冰壶到Ｇ时会立即停下

冰壶掉出区域算失败

求冰壶从Ｓ到Ｇ最少要推几次

如果不能在10次能完成输出-1,否则输出最小要推的次数

解题思路：

深度优先搜索

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

#define MAX_W 20
#define INF 99

int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};

int w,h;
int sx,sy;
int ex,ey;
int map[MAX_W][MAX_W];
int mintimes;

void dfs(int x,int y,int cur)
{
	if(cur==10 && map[y][x]!=3) return;
	else if(map[y][x]==3)
	{
		mintimes=min(mintimes,cur);
		return;
	}

	for(int i=0;i<4;i++)
	{
		int nx=x+dx[i],ny=y+dy[i];

		if(nx>=0 && nx<w && ny>=0 && ny<h)
		{
			if(map[ny][nx]==1)
				continue;

			while(nx>=0 && nx<w && ny>=0 && ny<h && map[ny][nx]!=1 && map[ny][nx]!=3)
			{
				nx+=dx[i];ny+=dy[i];
			}

			if(nx<0 || nx>=w || ny<0 || ny>=h) continue;

			int co=map[ny][nx];
			int cx=nx,cy=ny;
			if(map[ny][nx]==1)
			{
				map[ny][nx]=0;
				ny-=dy[i];nx-=dx[i];
			}
			dfs(nx,ny,cur+1);
			map[cy][cx]=co;
		}
	}
}

int main()
{
	scanf("%d%d",&w,&h);

	while(w!=0 || h!=0)
	{
		for(int i=0;i<h;i++)
		for(int j=0;j<w;j++)
		{
			scanf("%d",&map[i][j]);
			if(map[i][j]==2)
			{
				sx=j;sy=i;
			}
			else if(map[i][j]==3)
			{
				ex=j;ey=i;
			}
		}

		mintimes=INF;
		dfs(sx,sy,0);
		if(mintimes==INF)
			printf("-1\n");
		else
			printf("%d\n",mintimes);

		scanf("%d%d",&w,&h);
	}

	return 0;
}