Knights in FEN UVA10422

Output

For each set your task is to find the minimum number of moves leading from the starting input configuration to the final one. If that number is bigger than 10, then output one line stating

Unsolvable in less than 11 move(s).

otherwise output one line stating

Solvable in n move(s).

where n <= 10.

The output for each set is produced in a single line as shown in the sample output.

Sample Input

2

01011

110 1

01110

01010

00100

10110

01 11

10111

01001

00000

Sample Output

Unsolvable in less than 11 move(s).

Solvable in 7 move(s).

(Problem Setter: Piotr Rudnicki, University of Alberta, Canada)

“A man is as great as his dreams.”

題意：給出一個狀態圖，移動其中的騎士恢復到初始的狀態。

典型的隱式圖搜索問題，BFS搜索+哈希，哈希我是用的set實現的，另外要注意的是國際象棋中騎士走的是日字。

#include
#include
#include
#include

using namespace std;

typedef int state[25];
const int maxn=5000000;
const int dx[]={1,2,2,1,-1,-2,-2,-1};
const int dy[]={-2,-1,1,2,2,1,-1,-2};
state st[maxn];
int dist[maxn];
int front,rear,s;
set vis;
state goal={1,1,1,1,1,0,1,1,1,1,0,0,2,1,1,0,0,0,0,1,0,0,0,0,0};

int try_to_insert(int s)//哈希函數
{
    int v=0;
    for(int i=0;i<25;i++) v=v*2+st[s][i];
    if(vis.count(v))return 0;
    vis.insert(v);
    return 1;
}

int bfs()
{
    front=1,rear=2;
    vis.clear();
    while(front10) return -1;//剪枝，超過10步認為不可達
        if(memcmp(goal,p,sizeof(p))==0) return front;
        int i,j,x,y;
        for(i=0;i<25;i++) if(st[front][i]==2) break;
        int z=i;
        x=i/5,y=i%5;
        for(i=0;i<8;i++)
        {
            int newx=x+dx[i];
            int newy=y+dy[i];
            if(newx>=0&&newx<5&&newy>=0&&newy<5)
            {
                state &u=st[rear];
                memcpy(&u,&p,sizeof(p));
                u[x*5+y]=u[newx*5+newy];
                u[newx*5+newy]=2;
                dist[rear]=dist[front]+1;
                if(try_to_insert(rear)) rear++;
            }
        }
        front++;
    }
}

int main()
{
    cin>>s;
    string str;
    getline(cin,str);
    while(s--)
    {
        memset(dist,0,sizeof(dist));
        int i,j;
        for(i=0;i<5;i++)
        {
            getline(cin,str);
            for(j=0;j<5;j++)
            {
                if(str[j]!=' ') st[1][i*5+j]=str[j]-'0';
                else st[1][i*5+j]=2;
            }
        }
        int d=bfs();
        if(d<=0) cout<<"Unsolvable in less than 11 move(s)."<