hdu 1401 bfs

Solitaire

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1871 Accepted Submission(s): 629

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 52 4 3 3 3 6 4 6

Sample Output

YES

#include<iostream>#include<cstdio>#include<cstring>#include<queue>using  namespace std;/*尽管网上都是双广的解法，俺还是执着愚钝的单广...不小心还MLE了一次，于是vis改成了bool，且为88888888*/struct node{    int x[4],y[4],step;}p;int move[][2]={0,1,1,0,0,-1,-1,0};bool vis[8][8][8][8][8][8][8][8];bool mat[9][9];void set(node s){    vis[s.x[0]][s.y[0]][s.x[1]][s.y[1]][s.x[2]][s.y[2]][s.x[3]][s.y[3]]=1;}bool test(node s){    return vis[s.x[0]][s.y[0]][s.x[1]][s.y[1]][s.x[2]][s.y[2]][s.x[3]][s.y[3]];}bool okend(node s){    int i;    for(i=0;mat[s.x[i]][s.y[i]]&&i<4;++i);    return (i==4);}bool inboard(int x,int y){    if(x>=8||x<0||y>=8||y<0)        return 0;    return 1;}bool isempty(node s,int k){    for(int i=0;i<4;++i)        if(i!=k&&s.x[i]==s.x[k]&&s.y[i]==s.y[k])            return 0;    return 1;}bool bfs(){    p.step=0;    if(okend(p))        return 1;    memset(vis,0,sizeof(vis));    queue<node> que;    que.push(p);    int i,j,k;    node t;    while(!que.empty())    {        p=que.front();        /*好久以前一直没A的题，因为一些事就放下了.        之所以执着的些单广是因为结果是WA而不是TLE，终于找到原因        ，这里之前竟然是>8，不淡定啊，然后就AC了。        */        if(p.step>=8)            return 0;        que.pop();        for(k=0;k<4;++k)            for(i=0;i<4;++i)            {                t=p;                t.step++;                t.x[k]+=move[i][0];                t.y[k]+=move[i][1];                if(inboard(t.x[k],t.y[k]))                {                    if(!test(t))                    {                        if(isempty(t,k))                        {                            if(okend(t))                                return 1;                            set(t);                            que.push(t);                        }                        else                        {                            t.x[k]+=move[i][0];                            t.y[k]+=move[i][1];                            if(inboard(t.x[k],t.y[k])&&isempty(t,k)&&!test(t))                            {                                if(okend(t))                                    return 1;                                set(t);                                que.push(t);                            }                        }                    }                }            }    }    return 0;}int main(){    int i,x,y;    while(scanf("%d%d",&x,&y)==2)    {        --x;        --y;        p.x[0]=x;        p.y[0]=y;        for(i=1;i<4;++i)        {            scanf("%d%d",&x,&y);            --x;            --y;            p.x[i]=x;            p.y[i]=y;        }        memset(mat,0,sizeof(mat));        for(i=0;i<4;++i)        {            scanf("%d%d",&x,&y);            --x;            --y;            mat[x][y]=1;        }        printf(bfs()?"YES\n":"NO\n");    }    return 0;}