状态压缩
用01表示当前格子是否是正确的颜色，然后用十进制存储。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <queue> 7 using namespace std; 8 #define N 250007 9 10 struct node 11 { 12 int num,step; 13 }S; 14 15 char E[3][18]; 16 int vis[70000]; 17 int n,C; 18 19 int BFS(node S) 20 { 21 queue<node> que; 22 que.push(S); 23 vis[S.num] = 1; 24 int D = ((1<<(2*n))-1); 25 while(!que.empty()) 26 { 27 node tmp = que.front(); 28 que.pop(); 29 if(tmp.num == D) 30 return tmp.step; 31 int i,j,a,b; 32 for(i=0;i<2;i++) 33 { 34 for(j=0;j<n;j++) 35 { 36 int pos = i*n+j; 37 node now; 38 if(tmp.num & (1<<pos)) //已经是该颜色 39 continue; 40 int Num = 0; 41 for(a=pos;a<(i==0?n:2*n);a++) //往右 42 { 43 if(tmp.num & (1<<a)) //已经染好，不能再涂 44 break; 45 if(E[a/n][a%n] == E[pos/n][pos%n]) 46 Num = Num | (1<<a); //染色 47 } 48 for(a=pos-1;a>=(i==0?0:n);a--) //往左 49 { 50 if(tmp.num & (1<<a)) 51 break; 52 if(E[a/n][a%n] == E[pos/n][pos%n]) 53 Num = Num | (1<<a); 54 } 55 for(pos=Num;pos!=0;pos=Num&(pos-1)) //子状态 56 { 57 if(vis[tmp.num|pos]) 58 break; 59 vis[tmp.num|pos] = 1; 60 now.num = tmp.num|pos; 61 now.step = tmp.step + 1; 62 que.push(now); 63 } 64 pos = i*n+j; 65 if(i >= 1) 66 continue; 67 if(tmp.num & (1<<(n+pos))) 68 continue; 69 Num = 0; 70 for(a=pos;a<n;a++) //双行往右 71 { 72 if(tmp.num & (1<<a) || tmp.num & (1<<(a+n))) 73 break; 74 if(E[a/n][a%n] == E[pos/n][pos%n]) 75 Num = Num | (1<<a); 76 if(E[(a+n)/n][(a+n)%n] == E[pos/n][pos%n]) 77 Num = Num | (1<<(a+n)); 78 } 79 for(a=pos-1;a>=0;a--) //双行往左 80 { 81 if(tmp.num & (1<<a) || tmp.num & (1<<(a+n))) 82 break; 83 if(E[a/n][a%n] == E[pos/n][pos%n]) 84 Num = Num | (1<<a); 85 if(E[(a+n)/n][(a+n)%n] == E[pos/n][pos%n]) 86 Num = Num | (1<<(a+n)); 87 } 88 for(pos=Num;pos!=0;pos=Num&(pos-1)) //子状态 89 { 90 if(vis[tmp.num|pos]) 91 continue; 92 vis[tmp.num|pos] = 1; 93 now.num = tmp.num|pos; 94 now.step = tmp.step + 1; 95 que.push(now); 96 } 97 } 98 } 99 } 100 } 101 102 int main() 103 { 104 int t,cs = 1,i,j; 105 scanf("%d",&t); 106 while(t--) 107 { 108 scanf("%d",&n); 109 for(i=0;i<=1;i++) 110 scanf("%s",E[i]); 111 memset(vis,0,sizeof(vis)); 112 S.num = 0; 113 S.step = 0; 114 int res = BFS(S); 115 printf("Case #%d: %d ",cs++,res); 116 } 117 return 0; 118 }