uestc Can You Help God Wu

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\n",cs++,res);

116     }

117     return 0;

118 }