poj 3009 Curling 2.0

poj 3009

Curling 2.0

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 21445Accepted: 8742

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 square1 block2 start position3 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 13 26 61 0 0 2 1 01 1 0 0 0 00 0 0 0 0 30 0 0 0 0 01 0 0 0 0 10 1 1 1 1 16 11 1 2 1 1 36 11 0 2 1 1 312 12 0 1 1 1 1 1 1 1 1 1 313 12 0 1 1 1 1 1 1 1 1 1 1 30 0

Sample Output

14-1410-1

大致意思题目大意要求把一个冰壶从起点“2”用最少的步数移动到终点“3”，其中0为移动区域，1为石头区域，冰壶一旦想着某个方向运动就不会停止，也不会改变方向（想想冰壶在冰上滑动），除非冰壶撞到石头1或者到达终点3，冰壶撞到石头后，冰壶会停在石头前面，此时（静止状态）才允许改变冰壶的运动方向，而该块石头会破裂，石头所在的区域由1变为0. 也就是说，冰壶撞到石头后，并不会取代石头的位置。终点是一个摩擦力很大的区域，冰壶若到达终点3，就会停止在终点的位置不再移动。

#include<bits/stdc++.h>using namespace std;int square[5000][5000];int dir[4][2]={0,1,0,-1,-1,0,1,0};int n, m, ans;void DFS(int x, int y, int cnt){if(cnt>10)return; int i;for(i=0; i<4; i++){bool flag=1, tag=0;int nx=x, ny=y;while(1){nx+=dir[i][0];ny+=dir[i][1];if(nx<1 || nx>m || ny<1 || ny>n){flag=0;}if(square[nx][ny]==3 && ans > ++cnt){ans=cnt;return ;}if(square[nx][ny]==1 ){tag=1;}if(!flag)break;if(tag){square[nx][ny]=0;nx-=dir[i][0];ny-=dir[i][1];if(nx==x && ny==y){square[x+dir[i][0]][y+dir[i][1]]=1;break;}DFS(nx,ny,cnt+1);square[nx+dir[i][0]][ny+dir[i][1]]=1;break;}continue;}}}int main(){while(scanf("%d%d", &n,&m)==2){if(n==0)return 0;int sx, sy;ans=9999999;memset(square,0,sizeof(square));for(int i=1; i<=m; i++) for(int j=1; j<=n; j++){ scanf("%d", &square[i][j]); if(square[i][j]==2){sx=i;sy=j;}} DFS(sx,sy,0);if(ans<=10)printf("%d\n", ans);else puts("-1");}return 0;}

总结：

这是我好久没做acm题目后的第一道独立解决的问题，dfs，因为忘记清空数组查了好久，剩下的日子要快马加鞭了~~