银河之星（galaxy）

Posted 2020-08-30

当时没来的及做，应该是一道不是太难的搜索题。

首先，可以想到可行解的数量一定远远少于不可行解的数量；

与其我们直接搜索，倒不如我们根据（x,y）构造可行解，然后判断。

代码实现：

 1 #include<map>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 typedef long long LL;
 6 const int dx[8] = {0,1,0,-1,1,-1,1,-1};
 7 const int dy[8] = {1,0,-1,0,1,-1,-1,1};
 8 const int ch[9][9] = {{0,2,1,6,8,7,3,5,4},{2,1,0,8,7,6,5,4,3},{1,0,2,7,6,8,4,3,5},
 9                       {6,8,7,3,5,4,0,2,1},{8,7,6,5,4,3,2,1,0},{7,6,8,4,3,5,1,0,2},
10                       {3,5,4,0,2,1,6,8,7},{5,4,3,2,1,0,8,7,6},{4,3,5,1,0,2,7,6,8}};
11 map <LL,int> ma;
12 int n,m,k,X,Y,p[10][10],x[10],y[10];
13 LL now,goal,fac[10],che[10];
14 inline int ABS(int x) {return x < 0?-x:x;}
15 void Add(int X1,int Y1,int X2,int Y2){
16     int A=3*(X1%3)+Y1%3;
17     int B=3*(X2%3)+Y2%3;
18     p[A][B]=p[B][A]=1;
19 }
20 void Work(int a,int b){
21     for (int l=0;l<8;l++){
22         int xa=a+dx[l]*3;
23         int yb=b+dy[l]*3;
24         if(xa<1||xa>n||yb<1||yb>m) continue;
25         for(int i=0;i<8;i++){
26             int xx=xa+dx[i];
27             int yy=yb+dy[i];
28             if(xx<1||xx>n||yy<1||yy>m) continue;
29             Add(a,b,xx,yy);
30         }
31     }
32     for (int i=0;i<8;i++){
33         int xx=a+dx[i],yy=b+dy[i];
34            if(xx<1||xx>n||yy<1||yy>m) continue;
35            Add(a,b,xx,yy);
36     }
37 }
38 bool dfs(LL now,int tot){
39     if (now == goal) return 1;
40     if (ma[now]) return 0;
41     else ma[now] = 1;
42     if (tot == 1) return 0;
43     for (int i = 0; i < 9; i++)
44         for (int j = i + 1; j < 9; j++)
45             if (p[i][j] && che[i] && che[j]) {
46                 --che[i]; --che[j]; ++che[ch[i][j]];
47                 LL nex = 0;
48                 for (int l = 0; l < 9; l++) nex += fac[l]*che[l];
49                 if (dfs(nex,tot-1)) return 1;
50                 ++che[i]; ++che[j]; --che[ch[i][j]];
51             }
52     return 0;
53 }
54 int main(){
55     #ifdef YZY
56            freopen("galaxy4.in","r",stdin);
57     #endif
58     fac[0] = 1;
59     for (int i = 1; i < 10; i++) fac[i] = fac[i-1]*11LL;
60     while (scanf("%d%d%d%d%d",&k,&n,&m,&X,&Y) != EOF) {
61         memset(p,0,sizeof(p)); memset(che,0,sizeof(che)); ma.clear();
62         goal = fac[3*(X%3) + Y%3];
63         for (int i = 0; i < k; i++) {
64             scanf("%d%d",&x[i],&y[i]);
65             ++che[3*(x[i]%3) + y[i]%3];
66         }
67         for (int i = 1; i <= n; i++)
68             for (int j = 1; j <= m; j++)
69                 Work(i,j);
70         LL st = 0;
71         for (int i = 0; i < 9; i++) st += che[i]*fac[i];
72         if (dfs(st,k)) printf("Yes\n");
73         else printf("No\n");
74     }
75     return 0;
76 }

题目来源：Pku 3490