Solitaire
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4726 Accepted Submission(s): 1424
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
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
//八维数组记录每步的状态 的宽搜
using namespace std;
struct node{
int x[4],y[4];
int step;
}s,e;
bool vis[8][8][8][8][8][8][8][8];
bool map[10][10];
int to[4][2]={1,0,
-1,0,
0,1,
0,-1
};
int empty(node a,int k)
{
for(int i=0;i<4;i++)
{
if(i!=k&&a.x[i]==a.x[k]&&a.y[i]==a.y[k])
return 0;
}
return 1;
}
int judge(node tt)
{
for(int i=0;i<4;i++)
{
if(tt.x[i]<0||tt.x[i]>=8||tt.y[i]<0||tt.y[i]>=8)
return 1;
}
if(vis[tt.x[0]][tt.y[0]][tt.x[1]][tt.y[1]][tt.x[2]][tt.y[2]][tt.x[3]][tt.y[3]])
return 1;
return 0;
}
int check(node a)
{
for(int i=0;i<4;i++)
{
if(!map[a.x[i]][a.y[i]])
return 0;
}
return 1;
}
int bfs()
{
memset(vis,false,sizeof(vis));
queue<node> q;
node t,tt;
t.step=0;
for(int i=0;i<4;i++)
{
t.x[i]=s.x[i];
t.y[i]=s.y[i];
}
q.push(t);
vis[t.x[0]][t.y[0]][t.x[1]][t.y[1]][t.x[2]][t.y[2]][t.x[3]][t.y[3]]=1;
while(!q.empty())
{
t=q.front();
q.pop();
if(t.step>=8)
return 0;
if(check(t))
return 1;
for(int i=0;i<4;i++)
for(int j=0;j<4;j++)
{
tt=t;
tt.x[i]+=to[j][0];
tt.y[i]+=to[j][1];
tt.step++;
if(judge(tt))
continue;
if(empty(tt,i))
{
if(check(tt))
return 1;
vis[tt.x[0]][tt.y[0]][tt.x[1]][tt.y[1]][tt.x[2]][tt.y[2]][tt.x[3]][tt.y[3]]=true;
q.push(tt);
}
else
{
tt.x[i]+=to[j][0];
tt.y[i]+=to[j][1];
if(judge(tt)||!empty(tt,i)) continue;
if(check(tt)) return 1;
vis[tt.x[0]][tt.y[0]][tt.x[1]][tt.y[1]][tt.x[2]][tt.y[2]][tt.x[3]][tt.y[3]]=true;
q.push(tt);
}
}
}
return 0;
}
int main()
{
int i;
while(scanf("%d%d",&s.x[0],&s.y[0])!=EOF)
{
s.x[0]--;
s.y[0]--;
for(i=1;i<4;i++)
{
scanf("%d%d",&s.x[i],&s.y[i]);
s.x[i]--;
s.y[i]--;
}
memset(map,false,sizeof(map));
for(i=0;i<4;i++)
{
scanf("%d%d",&e.x[i],&e.y[i]);
e.x[i]--;
e.y[i]--;
map[e.x[i]][e.y[i]]=true;
}
int flag=bfs();
if(flag) puts("YES");
else puts("NO");
}
return 0;
}
