Solitaire(单向BFS||双向BFS)

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1401

Problem：

Solitaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3309    Accepted Submission(s): 1035

Problem Description

Solitaire is a game played on a chessboard 8x8. The rows and columns of the chessboard are numbered from 1 to 8, from the top to the bottom and from left to right respectively.

There are four identical pieces on the board. In one move it is allowed to:

> move a piece to an empty neighboring field (up, down, left or right),

> jump over one neighboring piece to an empty field (up, down, left or right). 



There are 4 moves allowed for each piece in the configuration shown above. As an example let's consider a piece placed in the row 4, column 4. It can be moved one row up, two rows down, one column left or two columns right.

Write a program that:

> reads two chessboard configurations from the standard input,

> verifies whether the second one is reachable from the first one in at most 8 moves,

> writes the result to the standard output.

 

Input

Each of two input lines contains 8 integers a1, a2, ..., a8 separated by single spaces and describes one configuration of pieces on the chessboard. Integers a2j-1 and a2j (1 <= j <= 4) describe the position of one piece - the row number and the column number respectively. Process to the end of file.

 

Output

The output should contain one word for each test case - YES if a configuration described in the second input line is reachable from the configuration described in the first input line in at most 8 moves, or one word NO otherwise.

 

Sample Input

  
  

   
   4 4 4 5 5 4 6 5
2 4 3 3 3 6 4 6
  
  

 

Sample Output

  
  

   
   YES
  
  

 

Source

Southwestern Europe 2002

 

题意就是有四颗棋子，能否在8步内走到目标状态

开个8维数组记录状态即可

 单向BFS：

[cpp]  view plain copy

1. #include <stdio.h>  
2. #include <string.h>  
3. #include <algorithm>  
4. #include <queue>  
5. using namespace std;  
6.   
7. bool vis[8][8][8][8][8][8][8][8];  
8. bool map[10][10];  
9. int to[4][2] = {1,0,-1,0,0,1,0,-1};  
10.   
11. struct point  
12. {  
13.     int x[4],y[4],step;  
14. } s,e;  
15.   
16. int check(point a)//判断是否为最终态  
17. {  
18.     for(int i = 0; i<4; i++)  
19.     {  
20.         if(!map[a.x[i]][a.y[i]])  
21.             return 0;  
22.     }  
23.     return 1;  
24. }  
25.   
26. int empty(point a,int k)//看要将要到达的那格是否为空  
27. {  
28.     for(int i = 0; i<4; i++)  
29.     {  
30.         if(i!=k && a.x[i] == a.x[k] && a.y[i] == a.y[k])  
31.             return 0;  
32.     }  
33.     return 1;  
34. }  
35.   
36. int judge(point next)//判断是否符合要求  
37. {  
38.     int i;  
39.     for(i = 0; i<4; i++)  
40.     {  
41.         if(next.x[i]<0 || next.x[i]>=8 || next.y[i]<0 || next.y[i]>=8)  
42.             return 1;  
43.     }  
44.     if(vis[next.x[0]][next.y[0]][next.x[1]][next.y[1]][next.x[2]][next.y[2]][next.x[3]][next.y[3]])  
45.         return 1;  
46.     return 0;  
47. }  
48.   
49. int bfs()  
50. {  
51.     memset(vis,false,sizeof(vis));  
52.     int i,j;  
53.     queue<point> Q;  
54.     point a,next;  
55.     a.step = 0;  
56.     for(i = 0; i<4; i++)  
57.     {  
58.         a.x[i] = s.x[i];  
59.         a.y[i] = s.y[i];  
60.     }  
61.     Q.push(a);  
62.     vis[a.x[0]][a.y[0]][a.x[1]][a.y[1]][a.x[2]][a.y[2]][a.x[3]][a.y[3]] = true;  
63.     while(!Q.empty())  
64.     {  
65.         a = Q.front();  
66.         Q.pop();  
67.         if(a.step>=8)//因为后面循环有判断减枝，所以这里要包括8步  
68.             return 0;  
69.         if(check(a))  
70.             return 1;  
71.         for(i = 0; i<4; i++)  
72.         {  
73.             for(j = 0; j<4; j++)  
74.             {  
75.                 next = a;  
76.                 next.x[i]+=to[j][0];  
77.                 next.y[i]+=to[j][1];  
78.                 next.step++;  
79.                 if(judge(next))  
80.                     continue;  
81.                 if(empty(next,i))//要去的那一格是空  
82.                 {  
83.                     if(check(next))  
84.                         return 1;  
85.                     vis[next.x[0]][next.y[0]][next.x[1]][next.y[1]][next.x[2]][next.y[2]][next.x[3]][next.y[3]] = true;  
86.                     Q.push(next);  
87.                 }  
88.                 else//非空则继续往前  
89.                 {  
90.                     next.x[i]+=to[j][0];  
91.                     next.y[i]+=to[j][1];  
92.                     if(judge(next) || !empty(next,i))//继续往前也要满足要求且是空格  
93.                         continue;  
94.                     if(check(next))  
95.                         return 1;  
96.                     vis[next.x[0]][next.y[0]][next.x[1]][next.y[1]][next.x[2]][next.y[2]][next.x[3]][next.y[3]] = true;  
97.                     Q.push(next);  
98.                 }  
99.             }  
100.         }  
101.     }  
102.     return 0;  
103. }  
104.   
105. int main()  
106. {  
107.     int i;  
108.     while(~scanf("%d%d",&s.x[0],&s.y[0]))  
109.     {  
110.         s.x[0]--;  
111.         s.y[0]--;  
112.         for(i = 1; i<4; i++)  
113.         {  
114.             scanf("%d%d",&s.x[i],&s.y[i]);  
115.             s.x[i]--;  
116.             s.y[i]--;  
117.         }  
118.         memset(map,false,sizeof(map));  
119.         for(i = 0; i<4; i++)  
120.         {  
121.             scanf("%d%d",&e.x[i],&e.y[i]);  
122.             e.x[i]--;  
123.             e.y[i]--;  
124.             map[e.x[i]][e.y[i]] = true;  
125.         }  
126.         int flag = bfs();  
127.         if(flag)  
128.             printf("YES\n");  
129.         else  
130.             printf("NO\n");  
131.     }  
132.   
133.     return 0;  
134. }