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 Input	Sample 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

 #include <iostream>

 #include <cstring>

 #include <cstdio>

 #include <algorithm>

 #include <queue>

 using namespace std;

 char g[];

 int  len,all;

 bool vis[];

 int  target;

 int  caculate[] = {,,,,,,,,,,,,,,,,};

 typedef struct status

 {

 int st,step,h;

 friend bool operator < (status a,status b)

 {

  if (a.step + a.h < b.step + b.h)

   return false;

  if (a.step + a.h == b.step + b.h && a.step  < b.step)

   return false;

  return true;

 }

 };

 priority_queue<status>q;

 int A(status &x)

 {

  bool ll[];

  int result = ;

  memset(ll,false,sizeof(ll));

  for(int i = ; i < all;++i)

   if ( (!(x.st & ( << i))) && !ll[g[i]-'A'])

    {

        result ++;

        ll[g[i]-'A'] = true;

    }

 return result;

 }

 int bfs()

 {

 status start;

 start.step = ,start.st = ;

 start.h = A(start);

 q.push(start);

 vis[] = true;

 while(!q.empty())

 {

  status ss = q.top();q.pop();

  if (ss.st == target) return ss.step;

  for(int i =  ; i <len;++i)

   for(int j = ;j <= len-i;++j)

    {

         char id;

         for(int h = ;h< j;++h)

          {

         id = g[i+h];

         int t = ss.st;

         for(int v =  ; v < j ;++ v)

          if(g[i+v] != id)

           t &= ~( << (i+v));

          else

           t |=  ( << (i+v));

         if (!vis[t])

          {

              status ns;

              ns.step = ss.step + ;

              ns.st = t;

              ns.h = A(ns);

              q.push(ns);

              vis[t] = true;

          }

       }

      for(int h = ;h< j;++h)

          {

         id = g[i+h+len];

         int t = ss.st;

         for(int v =  ; v < j ;++ v)

          if(g[i+v+len] != id)

           t &= ~( << (i+v+len));

          else

           t |=  ( << (i+v+len));

         if (!vis[t])

          {

              status ns;

              ns.step = ss.step + ;

              ns.h = A(ns);

              ns.st = t;

              q.push(ns);

              vis[t] = true;

          }

       }

      for(int h =  ; h < *j;++ h)

       {

           if (h >= j)

            id = g[h-j+i+len];

         else

          id = g[i+h];

         int t = ss.st;

         for(int v =  ; v < j ; ++ v)

          {

              if (g[i+v] != id)

               t &= ~( << (i+v));

                else

                 t |= ( << (i+v));

                if (g[i+v+len] != id)

                 t &= ~( << (i+v+len));

             else

              t |= ( << (i+v+len));

          }

          if (!vis[t])

           {

               vis[t] = true;

               status ns;

               ns.h = A(ns);

               ns.step  = ss.step + ;

               ns.st = t;

               q.push(ns);

           }

         } 

    }

 }

 return -;

 }

 int main(int argc, char * argv[])

 {

  int T,T2=;

  memset(g,,sizeof(g));

  scanf("%d",&T);

  while (T--)

   {

       while(!q.empty())

        q.pop();

       scanf("%d",&len);

       scanf("%s%s",g,&g[len]);

       memset(vis,false,sizeof(vis));

       target = caculate[len*]-;

       all = len*;

       cout << "Case #" << T2++ << ": " << bfs() << endl;

   }

  return ;

 }

巴特西

UESTC_Can You Help God Wu CDOJ 582

Input

Output

Sample input and output

Source

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