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。

/**
4 4 4 5 5 4 6 5
2 4 3 3 3 6 4 6

**/

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<algorithm>
#include<set>
using namespace std;

struct node
{
    int x[4];
    int y[4];
};
struct node start,end1;

queue<node>Q[2];
set<int>hxl[2];
bool flag;
int map1[4][2]={{1,0},{0,1},{-1,0},{0,-1}};

bool fun(node &t,int i)
{
    int j;
    if(t.x[i]>=1&&t.x[i]<=8  &&  t.y[i]>=1&&t.y[i]<=8)
    {
        for(j=0;j<4;j++)
        {
            if(j==i)continue;
            if(t.x[i]==t.x[j] && t.y[i]==t.y[j]) return true;
        }
        return false;
    }
    return true;
}
void pai(node &t)
{
   int i,j,x;
   for(i=0;i<4; i++)
   {
      x=i;
      for(j=i+1;j<4; j++)
       if(t.x[x]>t.x[j])
         x=j;
       else if(t.x[x]==t.x[j]  &&  t.y[x]>t.y[j])
         x=j;
      swap(t.x[i],t.x[x]);
      swap(t.y[i],t.y[x]);
   }

}
int serch(node &t)
{
    int i,sum=0;
    for(i=0;i<4;i++)
        sum=sum*100+t.x[i]*10+t.y[i];
    return sum;
}
void bfs(int x)
{
    int i,j,size1,k;
    node cur,t;
    size1=Q[x].size();
    while(size1--)
    {
        cur=Q[x].front();
        Q[x].pop();
        for(i=0;i<4;i++)/** every four point  **/
        {
            for(j=0;j<4;j++) /** n s w e**/
            {
                t=cur;
                t.x[i]=t.x[i]+map1[j][0];
                t.y[i]=t.y[i]+map1[j][1];
                if(fun(t,i)==true)
                {
                    t.x[i]=t.x[i]+map1[j][0];
                    t.y[i]=t.y[i]+map1[j][1];
                    if(fun(t,i)==true) continue;
                }
                pai(t);
                k=serch(t);
                if(hxl[x].count(k)>0)continue;
                if(hxl[x^1].count(k)>0)
                {
                    flag=true;
                    return;
                }
                hxl[x].insert(k);
                Q[x].push(t);
            }
        }
    }
}
void dbfs()
{
    int ans=0,k;
    pai(start);
    k=serch(start);
    hxl[0].insert(k);
    Q[0].push(start);

    pai(end1);
    Q[1].push(end1);
    k=serch(end1);
    hxl[1].insert(k);
    while(true)
    {
        if(Q[0].size()<Q[1].size())
            bfs(0);
        else bfs(1);
        ans++;
        if(ans==8)break;
        if(flag==true) return;
    }
}
int main()
{
    int i;
    while(scanf("%d%d",&start.x[0],&start.y[0])>0)
    {
        for(i=1;i<4;i++)
            scanf("%d%d",&start.x[i],&start.y[i]);
        for(i=0;i<4;i++)
            scanf("%d%d",&end1.x[i],&end1.y[i]);
        while(!Q[0].empty()){
            Q[0].pop();
        }
        while(!Q[1].empty()){
            Q[1].pop();
        }
        hxl[0].clear();
        hxl[1].clear();
        flag=false;
        dbfs();
        if(flag==true) printf("YES
");
        else printf("NO
");
    }
    return 0;
}