There is a boy named God Wu in UESTC ACM team. One day he is asked to finish a task. The task is that he has to paint a wall as the given pattern. The wall can be represented as a 2×n grid and each grid will be painted only one color. You know God Wu is a God, so he has a brush which could paint a rectangular area of the wall at a single step. As we all know, the paint could be overlap and the newly painted color will replace the older one.
God Wu is so lazy that he always want to finish something in the least steps. At first, the wall was not painted until God Wu paint some colors on it. For a given pattern of the wall, God Wu has to find out the minimum possible number of painting steps to make the wall the same as the given pattern.
Input
In the input file, the first line contains the number of test cases.
For each test case, the first contains only one integer n(1≤n≤8) indicating the length of the wall.
Then follows two lines, denoting the expected pattern of the wall. Every grid is painted by a color which is represented by a single capital letter. You can see the Input Sample for more details.
Output
For each test case, output only one integer denoting the minimum number of steps.
Sample input and output
| Sample Input | Sample Output |
|---|---|
3 3 ABA CBC 3 BAA CCB 3 BBB BAB |
Case #1: 3 Case #2: 3 Case #3: 2 |
Source
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <algorithm> 5 #include <queue> 6 using namespace std; 7 8 char g[18]; 9 int len,all; 10 bool vis[65537]; 11 int target; 12 int caculate[17] = {1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536}; 13 typedef struct status 14 { 15 int st,step,h; 16 friend bool operator < (status a,status b) 17 { 18 if (a.step + a.h < b.step + b.h) 19 return false; 20 if (a.step + a.h == b.step + b.h && a.step < b.step) 21 return false; 22 return true; 23 } 24 25 }; 26 27 priority_queue<status>q; 28 29 int A(status &x) 30 { 31 bool ll[26]; 32 int result = 0; 33 memset(ll,false,sizeof(ll)); 34 for(int i = 0; i < all;++i) 35 if ( (!(x.st & (1 << i))) && !ll[g[i]-'A']) 36 { 37 result ++; 38 ll[g[i]-'A'] = true; 39 } 40 return result; 41 } 42 43 44 int bfs() 45 { 46 status start; 47 start.step = 0,start.st = 0; 48 start.h = A(start); 49 q.push(start); 50 vis[0] = true; 51 while(!q.empty()) 52 { 53 status ss = q.top();q.pop(); 54 if (ss.st == target) return ss.step; 55 for(int i = 0 ; i <len;++i) 56 for(int j = 1;j <= len-i;++j) 57 { 58 char id; 59 for(int h = 0;h< j;++h) 60 { 61 id = g[i+h]; 62 int t = ss.st; 63 for(int v = 0 ; v < j ;++ v) 64 if(g[i+v] != id) 65 t &= ~(1 << (i+v)); 66 else 67 t |= (1 << (i+v)); 68 if (!vis[t]) 69 { 70 status ns; 71 ns.step = ss.step + 1; 72 ns.st = t; 73 ns.h = A(ns); 74 q.push(ns); 75 vis[t] = true; 76 } 77 } 78 79 for(int h = 0;h< j;++h) 80 { 81 id = g[i+h+len]; 82 int t = ss.st; 83 for(int v = 0 ; v < j ;++ v) 84 if(g[i+v+len] != id) 85 t &= ~(1 << (i+v+len)); 86 else 87 t |= (1 << (i+v+len)); 88 if (!vis[t]) 89 { 90 status ns; 91 ns.step = ss.step + 1; 92 ns.h = A(ns); 93 ns.st = t; 94 q.push(ns); 95 vis[t] = true; 96 } 97 } 98 99 100 101 for(int h = 0 ; h < 2*j;++ h) 102 { 103 if (h >= j) 104 id = g[h-j+i+len]; 105 else 106 id = g[i+h]; 107 int t = ss.st; 108 109 for(int v = 0 ; v < j ; ++ v) 110 { 111 if (g[i+v] != id) 112 t &= ~(1 << (i+v)); 113 else 114 t |= (1 << (i+v)); 115 116 if (g[i+v+len] != id) 117 t &= ~(1 << (i+v+len)); 118 else 119 t |= (1 << (i+v+len)); 120 121 } 122 123 if (!vis[t]) 124 { 125 vis[t] = true; 126 status ns; 127 ns.h = A(ns); 128 ns.step = ss.step + 1; 129 ns.st = t; 130 q.push(ns); 131 } 132 } 133 134 } 135 136 } 137 return -1; 138 } 139 140 141 int main(int argc, char * argv[]) 142 { 143 int T,T2=1; 144 memset(g,0,sizeof(g)); 145 scanf("%d",&T); 146 while (T--) 147 { 148 while(!q.empty()) 149 q.pop(); 150 scanf("%d",&len); 151 scanf("%s%s",g,&g[len]); 152 memset(vis,false,sizeof(vis)); 153 target = caculate[len*2]-1; 154 all = len*2; 155 cout << "Case #" << T2++ << ": " << bfs() << endl; 156 } 157 return 0; 158 }