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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.
The movement of the stone obeys the following rules:
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).
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
一个石头，问能不能大它从起点移动到终点。能则输出最小次数。不能则输出-1。
虽然要求是最小次数，但是这道题要用dfs、
要求：移动的次数<=10，
移动的规则：每次你可以选择4个方向移动，但是你一推石头，它会向着那个方向一直滚，直到经过目标或碰到block，
此时，石头停止，但是前面的block会被撞坏，成为空地。
或者一直滚，然后滚出了maze的范围，此时认为这个方向行不通。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 5 using namespace std; 6 7 const int maxn=103; 8 9 struct Point 10 { 11 int x,y; 12 }; 13 Point start,goal; 14 int row,col; 15 int ans; 16 17 int maze[maxn][maxn]; 18 19 int dx[4]={0,0,-1,1}; 20 int dy[4]={1,-1,0,0}; 21 22 bool isInRange(int x,int y) 23 { 24 if(x<1||x>row||y<1||y>col) 25 return false; 26 return true; 27 } 28 29 void dfs(Point cur,int dep) 30 { 31 if(dep>10) 32 return ; 33 34 for(int i=0;i<4;i++) 35 { 36 Point temp; 37 temp.x=cur.x+dx[i]; 38 temp.y=cur.y+dy[i]; 39 40 if(!isInRange(temp.x,temp.y)) 41 continue; 42 43 if(maze[temp.x][temp.y]) 44 continue; 45 46 if(temp.x==goal.x&&temp.y==goal.y) 47 { 48 ans=min(ans,dep+1); 49 return ; 50 } 51 52 while(isInRange(temp.x+dx[i],temp.y+dy[i])&& 53 !maze[temp.x+dx[i]][temp.y+dy[i]]) 54 { 55 temp.x+=dx[i]; 56 temp.y+=dy[i]; 57 if(temp.x==goal.x&&temp.y==goal.y) 58 { 59 ans=min(ans,dep+1); 60 return ; 61 } 62 } 63 64 if(!isInRange(temp.x+dx[i],temp.y+dy[i])) 65 continue; 66 67 else{ 68 maze[temp.x+dx[i]][temp.y+dy[i]]=0; 69 dfs(temp,dep+1); 70 maze[temp.x+dx[i]][temp.y+dy[i]]=1; 71 } 72 } 73 return ; 74 } 75 76 int main() 77 { 78 while(scanf("%d%d",&col,&row)) 79 { 80 if(row==0&&col==0) 81 break; 82 83 for(int i=1;i<=row;i++) 84 { 85 for(int j=1;j<=col;j++) 86 { 87 scanf("%d",&maze[i][j]); 88 if(maze[i][j]==2) 89 { 90 start.x=i; 91 start.y=j; 92 } 93 else if(maze[i][j]==3) 94 { 95 goal.x=i; 96 goal.y=j; 97 } 98 } 99 } 100 101 ans=11; 102 maze[start.x][start.y]=0; 103 maze[goal.x][goal.y]=0; 104 105 dfs(start,0); 106 107 if(ans<11) 108 { 109 printf("%d\n",ans); 110 } 111 else 112 { 113 printf("-1\n"); 114 } 115 } 116 117 return 0; 118 }
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原文地址：http://www.cnblogs.com/-maybe/p/4474921.html
