POJ3009 解题报告
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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.
- The stone hits a block (Fig. 2(b), (c)).
- 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.
0vacant square1block2start position3goal 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 0Sample Output
14-1410-1Source
#include<iostream>#include<algorithm>#include<set>using namespace std;const int maxw=22;const int maxh=22;int num[maxh][maxw];int dx[4]={-1,0,1,0};int dy[4]={0,1,0,-1};int sx,sy,gx,gy;int h,w;set<int>min_ans;void dfs(int x,int y,int ans);int main(){ while(cin>>w>>h) { if(w==0&&h==0) break; for(int i=0;i<h;i++) for(int j=0;j<w;j++) { cin>>num[i][j]; if(num[i][j]==2) { sx=i; sy=j; } if(num[i][j]==3) { gx=i; gy=j; } } min_ans.clear(); dfs(sx,sy,0); set<int>::iterator ite=min_ans.begin(); if(min_ans.size()) cout<<*ite<<endl; else cout<<-1<<endl; } return 0;}void dfs(int x,int y,int ans){ ans++; if(ans>10) return ; //如果步数大于10了，说明失败了，退出这一次调用 for(int i=0;i<4;i++) { int nx=x+dx[i],ny=y+dy[i]; //选择一个方向抛出冰壶 if(num[nx][ny]==1) continue; while( (0<=nx&&nx<h)&& //判断nx有没有出界 (0<=ny&&ny<w)&& //判断ny有没有出界 ( (num[nx][ny])!=3&&num[nx][ny]!=1 ) //判断是否到达终点或者是碰到石块 ) { nx+=dx[i]; //以上条件都不满足的时候就继续向当前方向前进 ny+=dy[i]; //以上条件都不满足的时候就继续向当前方向前进 } //循环结束，说明出现了上述情况，下面开始根据不同情况进行处理 if(num[nx][ny]==3) min_ans.insert(ans); //成功到达终点则将步数插入集合中 else if(num[nx][ny]==1) //碰到了石块，在石块的前一个位置停下来了 { num[nx][ny]=0; //由题意可知，碰到石块后，石块消失，于是将该位置设置为0 dfs(nx-dx[i],ny-dy[i],ans); //冰壶停下来了，再一次抛出冰壶，进行下一轮搜索 num[nx][ny]=1; //下一级搜索结束，这一级中该方向的搜索也结束了，于是将这个位置的石块恢复 } //不满足上面两个条件的话说明到了边界，不进行任何操作，直接换一个方向重新开始 }}#include<iostream>#include<algorithm>#include<set>using namespace std;const int maxw=22;const int maxh=22;int num[maxh][maxw]; //记录地图int dx[4]={-1,0,1,0};int dy[4]={0,1,0,-1}; //控制方向，下右上左int sx,sy,gx,gy; //s起点，g终点int h,w; //h行，w列set<int>min_ans; //存放到达终点时的步数void dfs(int x,int y,int ans);int main(){ while(cin>>w>>h) { if(w==0&&h==0) break; for(int i=0;i<h;i++) for(int j=0;j<w;j++) { cin>>num[i][j]; if(num[i][j]==2) { sx=i; sy=j; } if(num[i][j]==3) { gx=i; gy=j; } } min_ans.clear(); dfs(sx,sy,0); set<int>::iterator ite=min_ans.begin(); if(min_ans.size()) cout<<*ite<<endl; else cout<<-1<<endl; } return 0;}void dfs(int x,int y,int ans){ ans++; if(ans>10) return ; //如果步数大于10了，说明失败了，退出这一次调用 for(int i=0;i<4;i++) { int nx=x+dx[i],ny=y+dy[i]; //选择一个方向抛出冰壶 if(num[nx][ny]==1) continue; //如果这个方向就是石块，抛不出去，就换一个方向 while( (0<=nx&&nx<h)&& //判断nx有没有出界 (0<=ny&&ny<w)&& //判断ny有没有出界 ( (num[nx][ny])!=3&&num[nx][ny]!=1 ) //判断是否到达终点或者是碰到石块 ) { nx+=dx[i]; //以上条件都不满足的时候就继续向当前方向前进 ny+=dy[i]; //以上条件都不满足的时候就继续向当前方向前进 } //循环结束，说明冰壶碰到了石块、到达终点、出界 if((！(0<=nx&&nx<h)&&(0<=ny&&ny<w))) continue; else if(num[nx][ny]==3) min_ans.insert(ans); //成功到达终点则将步数插入集合中 else if(num[nx][ny]==1) //碰到了石块，在石块的前一个位置停下来了 { num[nx][ny]=0; //由题意可知，碰到石块后，石块消失，于是将该位置设置为0 dfs(nx-dx[i],ny-dy[i],ans); //冰壶停下来了，再一次抛出冰壶，进行下一轮搜索 num[nx][ny]=1; //下一级搜索结束，这一级中该方向的搜索也结束了，于是将这个位置的石块恢复 } //不满足上面两个条件的话说明到了边界，不进行任何操作，直接换一个方向重新开始 }}#include<iostream>#include<algorithm>#include<set>using namespace std;const int maxw=22;const int maxh=22;int num[maxh][maxw]; //记录地图int dx[4]={-1,0,1,0};int dy[4]={0,1,0,-1}; //控制方向，下右上左int sx,sy,gx,gy; //s起点，g终点int h,w; //h行，w列set<int>min_ans; //存放到达终点时的步数void dfs(int x,int y,int ans);int main(){ while(cin>>w>>h) { if(w==0&&h==0) break; for(int i=0;i<h;i++) for(int j=0;j<w;j++) { cin>>num[i][j]; if(num[i][j]==2) { sx=i; sy=j; } if(num[i][j]==3) { gx=i; gy=j; } } min_ans.clear(); dfs(sx,sy,0); set<int>::iterator ite=min_ans.begin(); if(min_ans.size()) cout<<*ite<<endl; else cout<<-1<<endl; } return 0;}void dfs(int x,int y,int ans){ ans++; if(ans>10) return ; //如果步数大于10了，说明失败了，退出这一次调用 for(int i=0;i<4;i++) { int nx=x+dx[i],ny=y+dy[i]; //选择一个方向抛出冰壶 if(num[nx][ny]==1) continue; //如果这个方向就是石块，抛不出去，就换一个方向 while( (0<=nx&&nx<h)&& //判断nx有没有出界 (0<=ny&&ny<w)&& //判断ny有没有出界 ( (num[nx][ny])!=3&&num[nx][ny]!=1 ) //判断是否到达终点或者是碰到石块 ) { nx+=dx[i]; //以上条件都不满足的时候就继续向当前方向前进 ny+=dy[i]; //以上条件都不满足的时候就继续向当前方向前进 } //循环结束，说明冰壶碰到了石块、到达终点、出界 if(!((0<=nx&&nx<h)&&(0<=ny&&ny<w))) continue; else if(num[nx][ny]==3) min_ans.insert(ans); //成功到达终点则将步数插入集合中 else if(num[nx][ny]==1) //碰到了石块，在石块的前一个位置停下来了 { num[nx][ny]=0; //由题意可知，碰到石块后，石块消失，于是将该位置设置为0 dfs(nx-dx[i],ny-dy[i],ans); //冰壶停下来了，再一次抛出冰壶，进行下一轮搜索 num[nx][ny]=1; //下一级搜索结束，这一级中该方向的搜索也结束了，于是将这个位置的石块恢复 } //不满足上面两个条件的话说明到了边界，不进行任何操作，直接换一个方向重新开始 }}最后下面是我的测试数据
3 91 1 13 0 01 0 01 0 01 0 01 0 02 0 10 0 01 0 15 11 2 1 1 33 91 1 13 0 01 0 01 0 01 0 01 0 02 0 10 0 01 1 13 61 1 13 0 01 0 02 0 10 0 01 1 12 12 33 12 1 3
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