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  • UESTC_Can You Help God Wu CDOJ 582

    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 InputSample Output
     
    3
3
ABA
CBC
3
BAA
CCB
3
BBB
BAB
    		 
    Case #1: 3
Case #2: 3
Case #3: 2

    Source

    UESTC Training for Search Algorithm
     
    解题报告
    暴力枚举涂法，注意二进制的优先级。。TLE
    改进：
    我们设计一个A*算法，g(x) = 目前的涂色次数 , h(x) = 当前的不对颜色种类.
    很容易得到 f(x) = g(x) + h(x)是满足单调性的
    证明：
     设在某步时g(n) = x,h(n) = y;
     在下一步时必然有 g(n') = x + 1 , 且h(n') = y or y-1, 因为你每次只能消除一种颜色，甚至于等于没消
     故 c(n,n') + h(n') - h(n) = 1 + (y or y-1) -y -> min = 0 >=0
     满足相容性，得证
      
      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 }
    No Pain , No Gain.
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  • 原文地址：https://www.cnblogs.com/Xiper/p/4455122.html
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