HDU - 1401 Solitaire

Solitaire

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

 

解题思路：简单的bfs，难点在于状态判断，这里使用了八维数组，刚好不会超空间。即把4个棋子的坐标作为状态，在这之前要先对棋子排序，因为1 1 2 2 3 3 4 4和4 4 3 3 2 2 1 1 其实是一样的状态。之后就是简单的bfs了。另外这题可以用双向bfs，即两边同时bfs，如果在4步内，状态能达到一模一样，那证明可以在8步内达到，这样可以大大优化时间。这里用的是单向bfs，直接搜到8步。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <vector>
#include <map>
#include <string.h>
#include <algorithm>
#include <queue>

using namespace std;

bool vis[8][8][8][8][8][8][8][8];

int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};

struct point{
int x[4],y[4];
int t;

point(int a1=0,int a2=0,int a3=0,int a4=0,int a5=0,int a6=0,int a7=0,int a8=0,int a=0){
x[0]=a1;
y[0]=a2;
x[1]=a3;
y[1]=a4;
x[2]=a5;
y[2]=a6;
x[3]=a7;
y[3]=a8;
t=a;
adjust();
}

void adjust(){
for(int i=0;i<4;i++)
for(int j=i+1;j<4;j++){
if(x[i]>x[j])
{
swap(x[i],x[j]);
swap(y[i],y[j]);
}
else{
if(x[i]==x[j]){
if(y[i]>y[j]){
swap(x[i],x[j]);
swap(y[i],y[j]);
}
}
}
}
}

};

point endp;
point start;

//判断两个点是否相等
bool same(point a,point b){
a.adjust();
b.adjust();

for(int i=0;i<4;i++){
if(a.x[i]!=b.x[i])
return false;
if(a.y[i]!=b.y[i])
return false;
}
return true;
}

//判断是否要走两步，即该格子有没有棋子
bool have(int x,int y,point a){
for(int i=0;i<4;i++)
if(a.x[i]==x&&a.y[i]==y)
return true;
return false;
}

bool bfs(){

queue<point> que;
que.push(start);

vis[start.x[0]-1][start.y[0]-1][start.x[1]-1][start.y[1]-1][start.x[2]-1][start.y[2]-1][start.x[3]-1][start.y[3]-1]=1;

while(!que.empty()){

point tp=que.front();
que.pop();

if(tp.t>8)
return false;

if(same(tp,endp))
return true;

int x,y;
//4个棋子，4个方向
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
x=tp.x[i]+dx[j];
y=tp.y[i]+dy[j];

if(x==0||y==0||x==9||y==9)
continue;

if(have(x,y,tp)){

if(j==0)
x--;
if(j==1)
x++;
if(j==2)
y--;
if(j==3)
y++;

if(x==0||y==0||x==9||y==9)
continue;

if(!have(x,y,tp)){
if(i==0){
point temp=point(x,y,tp.x[1],tp.y[1],tp.x[2],tp.y[2],tp.x[3],tp.y[3],tp.t+1);

if(vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]==0){
vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]=1;
que.push(temp);
}

}
if(i==1){
point temp=point(tp.x[0],tp.y[0],x,y,tp.x[2],tp.y[2],tp.x[3],tp.y[3],tp.t+1);

if(vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]==0){
vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]=1;
que.push(temp);
}

}
if(i==2){
point temp=point(tp.x[0],tp.y[0],tp.x[1],tp.y[1],x,y,tp.x[3],tp.y[3],tp.t+1);

if(vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]==0){
vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]=1;
que.push(temp);
}

}
if(i==3){
point temp=point(tp.x[0],tp.y[0],tp.x[1],tp.y[1],tp.x[2],tp.y[2],x,y,tp.t+1);

if(vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]==0){
vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]=1;
que.push(temp);
}

}
}
}
else{
if(i==0){
point temp=point(x,y,tp.x[1],tp.y[1],tp.x[2],tp.y[2],tp.x[3],tp.y[3],tp.t+1);

if(vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]==0){
vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]=1;
que.push(temp);
}

}
if(i==1){
point temp=point(tp.x[0],tp.y[0],x,y,tp.x[2],tp.y[2],tp.x[3],tp.y[3],tp.t+1);

if(vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]==0){
vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]=1;
que.push(temp);
}

}
if(i==2){
point temp=point(tp.x[0],tp.y[0],tp.x[1],tp.y[1],x,y,tp.x[3],tp.y[3],tp.t+1);

if(vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]==0){
vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]=1;
que.push(temp);
}
}
if(i==3){
point temp=point(tp.x[0],tp.y[0],tp.x[1],tp.y[1],tp.x[2],tp.y[2],x,y,tp.t+1);

if(vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]==0){
vis[temp.x[0]-1][temp.y[0]-1][temp.x[1]-1][temp.y[1]-1][temp.x[2]-1][temp.y[2]-1][temp.x[3]-1][temp.y[3]-1]=1;
que.push(temp);
}

}
}

}

}

}

return false;
}

int main()
{

int a1=0, a2=0, a3=0, a4=0, a5=0, a6=0, a7=0, a8=0;
while(cin>>a1>>a2>>a3>>a4>>a5>>a6>>a7>>a8){
start=point(a1,a2,a3,a4,a5,a6,a7,a8,0);
cin>>a1>>a2>>a3>>a4>>a5>>a6>>a7>>a8;
endp=point(a1,a2,a3,a4,a5,a6,a7,a8,0);

memset(vis,0,sizeof(vis));

if(bfs()){
cout<<"YES"<<endl;
}
else
cout<<"NO"<<endl;

}

return 0;
}