这个速度比分步快一点,内存占的稍微多一点
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
1 #include <iostream> 2 #include <algorithm> 3 #include <queue> 4 #include <map> 5 using namespace std; 6 7 const int dir[4][2] = {1,0,-1,0,0,1,0,-1}; 8 9 struct point 10 { 11 int x,y; 12 }; 13 struct node 14 { 15 point chess[4]; 16 17 bool check(int j) 18 { 19 if(chess[j].x>=0 && chess[j].x<=7 && chess[j].y>=0 && chess[j].y<=7) 20 { 21 for(int i=0;i<4;i++) 22 { 23 if(i!=j && chess[i].x==chess[j].x && chess[i].y==chess[j].y) 24 { 25 return false; 26 } 27 } 28 return true; 29 } 30 return false; 31 } 32 }s,e; 33 34 bool cmp(const struct point& a,const struct point& b) 35 { 36 if(a.x == b.x) 37 { 38 return a.y<b.y; 39 } 40 return a.x<b.x; 41 } 42 int gethash(node& a) 43 { 44 sort(a.chess, a.chess+4, cmp); 45 int hash = 0; 46 for(int i=0; i<4; i++) 47 { 48 hash |= a.chess[i].x << (6*i); 49 hash |= a.chess[i].y << (6*i+3); 50 } 51 return hash; 52 } 53 54 map<int,int>mapint; 55 map<int,int>::iterator it_1,it_2; 56 57 bool BFS(void) 58 { 59 queue<node>que[2]; 60 61 que[0].push(s); 62 que[1].push(e); 63 mapint[gethash(s)] = 10*0+0; 64 mapint[gethash(e)] = 10*1+0; 65 66 int sign; 67 while(!que[0].empty() || !que[1].empty()) 68 { 69 if(que[0].size() < que[1].size()) 70 { 71 sign = que[0].empty() ? 1:0; 72 } 73 else 74 { 75 sign = que[1].empty() ? 0:1; 76 } 77 78 node temp = que[sign].front(); 79 que[sign].pop(); 80 81 it_1 = mapint.find(gethash(temp)); 82 83 if((it_1->second)%10 >= 4) //移动步数超过4步 84 { 85 continue; 86 } 87 for(int i=0;i<4;i++) 88 { 89 for(int j=0;j<4;j++) 90 { 91 node next = temp; 92 next.chess[i].x += dir[j][0]; 93 next.chess[i].y += dir[j][1]; 94 95 if(!next.check(i)) //重叠或者越界 96 { 97 next.chess[i].x += dir[j][0]; 98 next.chess[i].y += dir[j][1]; 99 if(!next.check(i)) //重叠或者越界 100 { 101 continue; 102 } 103 } 104 105 int hash = gethash(next); 106 it_2 = mapint.find(hash); 107 108 if(it_2 == mapint.end()) 109 { 110 mapint[hash] = it_1->second + 1; 111 que[sign].push(next); 112 } 113 else if(it_2->second/10 == 1-sign) 114 { 115 return true; 116 } 117 } 118 } 119 } 120 return false; 121 } 122 123 int main() 124 { 125 while(cin>>s.chess[0].x>>s.chess[0].y) 126 { 127 for(int i=1; i<4; i++) 128 { 129 cin>>s.chess[i].x>>s.chess[i].y; 130 } 131 for(int i=0; i<4; i++) 132 { 133 cin>>e.chess[i].x>>e.chess[i].y; 134 } 135 for(int i=0;i<4;i++) 136 { 137 s.chess[i].x--; s.chess[i].y--; 138 e.chess[i].x--; e.chess[i].y--; 139 } 140 141 if(BFS()) 142 { 143 cout<<"YES"<<endl; 144 } 145 else 146 { 147 cout<<"NO"<<endl; 148 } 149 mapint.clear(); 150 } 151 return 0; 152 }