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
");
143         else printf("NO
");
144     }
145     return 0;
146 }