hdu 1401

Solitaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3077    Accepted Submission(s): 954

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

 

 

Recommend

四个点的哈希，由于这四个点没有什么区分，哈希的时候注意一下。

由于的用set哈希，用*10的方法。所以要先排序。很容易理解的。

bfs的代码，有点意思。

  1 /**

  2 4 4 4 5 5 4 6 5

  3 2 4 3 3 3 6 4 6

  4 

  5 **/

  6 

  7 #include<iostream>

  8 #include<stdio.h>

  9 #include<cstring>

 10 #include<cstdlib>

 11 #include<queue>

 12 #include<algorithm>

 13 #include<set>

 14 using namespace std;

 15 

 16 struct node

 17 {

 18     int x[4];

 19     int y[4];

 20 };

 21 struct node start,end1;

 22 

 23 queue<node>Q[2];

 24 set<int>hxl[2];

 25 bool flag;

 26 int map1[4][2]={{1,0},{0,1},{-1,0},{0,-1}};

 27 

 28 bool fun(node &t,int i)

 29 {

 30     int j;

 31     if(t.x[i]>=1&&t.x[i]<=8  &&  t.y[i]>=1&&t.y[i]<=8)

 32     {

 33         for(j=0;j<4;j++)

 34         {

 35             if(j==i)continue;

 36             if(t.x[i]==t.x[j] && t.y[i]==t.y[j]) return true;

 37         }

 38         return false;

 39     }

 40     return true;

 41 }

 42 void pai(node &t)

 43 {

 44    int i,j,x;

 45    for(i=0;i<4; i++)

 46    {

 47       x=i;

 48       for(j=i+1;j<4; j++)

 49        if(t.x[x]>t.x[j])

 50          x=j;

 51        else if(t.x[x]==t.x[j]  &&  t.y[x]>t.y[j])

 52          x=j;

 53       swap(t.x[i],t.x[x]);

 54       swap(t.y[i],t.y[x]);

 55    }

 56 

 57 }

 58 int serch(node &t)

 59 {

 60     int i,sum=0;

 61     for(i=0;i<4;i++)

 62         sum=sum*100+t.x[i]*10+t.y[i];

 63     return sum;

 64 }

 65 void bfs(int x)

 66 {

 67     int i,j,size1,k;

 68     node cur,t;

 69     size1=Q[x].size();

 70     while(size1--)

 71     {

 72         cur=Q[x].front();

 73         Q[x].pop();

 74         for(i=0;i<4;i++)/** every four point  **/

 75         {

 76             for(j=0;j<4;j++) /** n s w e**/

 77             {

 78                 t=cur;

 79                 t.x[i]=t.x[i]+map1[j][0];

 80                 t.y[i]=t.y[i]+map1[j][1];

 81                 if(fun(t,i)==true)

 82                 {

 83                     t.x[i]=t.x[i]+map1[j][0];

 84                     t.y[i]=t.y[i]+map1[j][1];

 85                     if(fun(t,i)==true) continue;

 86                 }

 87                 pai(t);

 88                 k=serch(t);

 89                 if(hxl[x].count(k)>0)continue;

 90                 if(hxl[x^1].count(k)>0)

 91                 {

 92                     flag=true;

 93                     return;

 94                 }

 95                 hxl[x].insert(k);

 96                 Q[x].push(t);

 97             }

 98         }

 99     }

100 }

101 void dbfs()

102 {

103     int ans=0,k;

104     pai(start);

105     k=serch(start);

106     hxl[0].insert(k);

107     Q[0].push(start);

108 

109     pai(end1);

110     Q[1].push(end1);

111     k=serch(end1);

112     hxl[1].insert(k);

113     while(true)

114     {

115         if(Q[0].size()<Q[1].size())

116             bfs(0);

117         else bfs(1);

118         ans++;

119         if(ans==8)break;

120         if(flag==true) return;

121     }

122 }

123 int main()

124 {

125     int i;

126     while(scanf("%d%d",&start.x[0],&start.y[0])>0)

127     {

128         for(i=1;i<4;i++)

129             scanf("%d%d",&start.x[i],&start.y[i]);

130         for(i=0;i<4;i++)

131             scanf("%d%d",&end1.x[i],&end1.y[i]);

132         while(!Q[0].empty()){

133             Q[0].pop();

134         }

135         while(!Q[1].empty()){

136             Q[1].pop();

137         }

138         hxl[0].clear();

139         hxl[1].clear();

140         flag=false;

141         dbfs();

142         if(flag==true) printf("YES\n");

143         else printf("NO\n");

144     }

145     return 0;

146 }