HDU 2514 Another Eight Puzzle

题目：http://acm.hdu.edu.cn/showproblem.php?pid=2514

对8个位置填数,有连线的两个位置不能填连续的数.

搜索时加上限定条件即可.除了搜索,记得还有一种更优雅的解法.

#include <iostream>
using namespace std;
int judge[8][8] = {{3,1,2,3},{4,0,2,4,5},{6,0,1,3,4,5,6},{4,0,2,5,6},{4,1,2,5,7},{6,1,2,3,4,6,7},{4,2,3,5,7},{3,4,5,6}};
//judge[i][0] = 第i列相连点的数量,judge[i][j](j>0) = 相连点.图的数组结构
int num = 0;     //记录解的数量
bool vis[9];     // vis记录8个数中被用过的数
int a[8],fil[8],seq[8];; // fil记录有多少个待填点
int n;
void dfs(int t)
{
    if(num > 1) return;
    if(t == n)
    {
        num ++ ;
        for(int i=0;i < 8;i++)
        {
            seq[i] = a[i];
        }
        return;
    }

    for(int i = 1;i <= 8;i++) //枚举8个数
    {
        if(!vis[i])
        {
            for(int j=1;j < judge[fil[t]][0];j++) //枚举相连点
            {
                if(abs(a[judge[fil[t]][j]] - i) == 1 && a[judge[fil[t]][j]] != 0) //若连续则跳过
                {
                    break;
                }
            }
            if(j < judge[fil[t]][0]) continue; 
            //若非正常退出,则出现连续break跳出则不满足条件
            //否则,进行深搜
            vis[i] = 1;
            a[fil[t]] = i;
            dfs(t+1);
            if(num > 1) return;
            vis[i] = 0;
            a[fil[t]] = 0;

        }
    }
}

int main(int argc, const char *argv[])
{
    int T;
//    freopen("input.txt","r",stdin);
    cin>>T;
    for(int k = 1;k <= T;k++)
    {
        n = 0;
        memset(vis,0,sizeof(vis));
        for(int i=0;i < 8;i++)
        {
            cin>>a[i];
            if(a[i]) vis[a[i]] = 1;
            else if(!a[i])  //若为0.fil记录位置
            {
                fil[n++] = i;
            }
        }
        num = 0;
        dfs(0);
        cout<<"Case "<<k<<":"<<" ";
        if(num == 1)
        {
            cout<<seq[0];
            for(int i=1;i < 8;i++)
            {
                cout<<" "<<seq[i];
            }
            cout<<endl;
        }
        else if(num > 1)
            cout<<"Not unique"<<endl;
        else cout<<"No answer"<<endl;
    }


    return 0;
}