HOJ---11491 A Knight and a Queen[超大数组+BFS]

 

A Knight and a Queen

Time Limit: 5000ms, Special Time Limit:12500ms, Memory Limit:32768KB

Total submit users: 52, Accepted users: 42

Problem 11491 : No special judgement

Problem description

Marge walks in the house and finds Homer staring at a chessboard with a knight and a queen on it. Marge: Homer, are you OK? What’s up with these intellectual endeavors of yours recently? Homer: Oh, I’m OK. And I’m not playing chess anyway. Marge: So what are you doing? Homer: I’m imagining that I’m a knight. Marge: Yeah right. Homer: Well, a knight sitting on the chessboard. And I imagine you are my queen. Marge: I like that. Homer: Of course you are sitting on the chessboard too. Now I wonder if I can reach the square you’re sitting on in 16 or fewer moves. Marge: So can you? Homer: I’m still trying to figure it out.   Write a program that, when given a knight and a queen on an n by n chessboard, finds if the knight can reach the queen in m or fewer than moves. One ”move” for a knight is defined as 2 squares in one direction, then one square in a perpendicular direction. Knights cannot move along diagonals. For example, if n = 8, kx = 3, and ky = 5, then possible positions for the knight after one move given in (kx, ky) are: (1,6), (1,4), (2,7), (2,3), (4,7), (4,3), (5,6), and (5,4).

Input

For each test the input consists of 3 lines. The first line contains 2 numbers: n and m. n is the dimension of the board, and m is the number of moves the knight is allowed to do.   The second line contains two numbers: kx and ky, s.t. 1 ≤ kx ≤ n, 1 ≤ ky ≤ n. These two numbers indicate the position of the knight on the board.   The last line again contains two numbers: qx and qy, s.t. 1 ≤ qx ≤ n, 1 ≤ qy ≤ n. They indicate the position of the queen on the board.

Output

If the queen is reachable by the knight with m or less moves, output should contain the following string on a single line ending with a newline: Knight can reach Queen within m moves!   If the queen is not reachable by the knight with m or less moves, output should contain the following string on a single line ending with a newline: Knight cannot reach Queen within m moves!   In the above, m should be the actual value from the input.   No test case will have the queen and the knight sitting on the same square. You can further assume that n ≤ 1000000,m ≤ 256. Your program should finish in less than one minute.

Sample Input

8 2 3 5 7 4 8 3 3 5 7 4

Sample Output

Knight cannot reach Queen within 2 moves! Knight can reach Queen within 3 moves!

Problem Source

2008 Maryland High-school Programming Contest

 

 

 

 

因为数组大小很大，所以要把起点和终点平移，然后注意边界就好。

code:

  1 /*
  2 16 9
  3 15 2
  4 2 15
  5 Knight cannot reach Queen within 9 moves!
  6 16 10
  7 15 2
  8 2 15
  9 Knight can reach Queen within 10 moves!
 10 8 3
 11 4 2
 12 4 8
 13 Knight cannot reach Queen within 3 moves!
 14 1000000 256
 15 200000 800000
 16 800000 200000
 17 Knight cannot reach Queen within 256 moves!
 18 1000000 256
 19 350000 350000
 20 650000 650000
 21 Knight cannot reach Queen within 256 moves!
 22 384 256
 23 1 1
 24 384 384
 25 Knight can reach Queen within 256 moves!
 26 96 64
 27 1 1
 28 96 96
 29 Knight can reach Queen within 96 moves!
 30 */
 31 
 32 #include    
 33 #include    
 34 #include    
 35 #include    
 36 #include    
 37 #include <string>   
 38 #include <set>   
 39 #include    
 40 #include    
 41 #include    
 42 #include    
 43 #include    
 44 #include    
 45 #include    
 46 #include    
 47 #include    
 48 #include    
 49 #include  
 50 using namespace std;
 51 
 52 int map[2000][2000];
 53 int vst[2000][2000];
 54 int n,m;
 55 int sx,sy,ex,ey;
 56 bool flag;
 57 int ledgex1,ledgey1,ledgex2,ledgey2;
 58 int dir[8][2]={
 59     2,1,
 60     1,2,
 61     -1,2,
 62     -2,1,
 63     -2,-1,
 64     -1,-2,
 65     1,-2,
 66     2,-1
 67 };
 68 
 69 struct node
 70 {
 71     int x,y;
 72     int step;
 73 }Node;
 74 
 75 bool check(int x,int y)
 76 {
 77     if(x>=(1000-ledgex1)&&x<(1000+n-ledgex1)&&y>=(1000-ledgey1)&&y<(1000+n-ledgey1)&&!vst[x][y])
 78         return true;
 79     else
 80         return false;
 81 }
 82 
 83 void bfs()
 84 {
 85     int i;
 86     node pre,last;
 87     pre.x=1000;
 88     pre.y=1000;
 89     pre.step=0;
 90     queueQue;
 91     Que.push(pre);
 92     while(!Que.empty())
 93     {
 94         pre=Que.front();
 95         Que.pop();
 96         if(pre.step>m)
 97         {
 98             flag=false;
 99              return;
100         }
101         if(pre.x==ex&&pre.y==ey)
102         {
103             if(pre.step>m)
104                 flag=false;
105             return;
106         }
107         for(i=0;i<8;i++)
108         {
109             last.x=pre.x+dir[i][0];
110             last.y=pre.y+dir[i][1];
111             last.step=pre.step+1;
112             if(check(last.x,last.y))
113             {
114                 vst[last.x][last.y]=1;
115                 Que.push(last);
116             }
117         }
118     }
119 }
120 
121 int main()
122 {
123     while(~scanf("%d%d%d%d%d%d",&n,&m,&sx,&sy,&ex,&ey))
124     {
125         flag=true;
126         memset(vst,0,sizeof(vst));
127         if(abs(ex-sx)>m*2||abs(ey-sy)>m*2)
128             flag=false;
129         else
130         {
131             sx--;
132             ex--;
133             sy--;
134             ey--;
135             ledgex1=(sx>ex?ex:sx)>513?513:(sx>ex?ex:sx);
136             ledgey1=(sy>ey?ey:sy)>513?513:(sy>ey?ey:sy);
137             ex=ex-sx+1000;
138             ey=ey-sy+1000;
139             bfs();
140         }
141         if(flag)
142             printf("Knight can reach Queen within %d moves!\n",m);
143         else
144             printf("Knight cannot reach Queen within %d moves!\n",m);
145     }
146     return 0;
147 }

 

 

 

学长的代码，用了#include

code2：

 1 #include 
 2 #include 
 3 #include