HDU 4012Paint on a Wall

Paint on a Wall

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 65768/65768K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

Problem Description

Annie wants to paint her wall to an expected pattern. The wall can be represented as a 2*n grid and each grid will be painted only one color. Annie has a brush which could paint a rectangular area of the wall at a single step. The paint could be overlap and the newly painted color will replace the older one.
  For a given pattern of the wall, your task is to help Annie find out the minimum possible number of painting steps. You can assume that the wall remains unpainted until Annie paint some colors on it.

 

Input

There are multiple test cases in the input. The first line contains the number of test cases. For each test case, the first line contains the only integer n indicating the length of the wall. (1 <= n <= 8) Two lines follow, denoting the expected pattern of the wall. The color is represented by a single capital letter. See the sample input for further details.

 

Output

For each test case, output only one integer denoting the minimum number of steps.

 

Sample Input

   
   
   
   

    
    
    
    3
3
ABA
CBC
3
BAA
CCB
3
BBB
BAB

   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 3
Case #2: 3
Case #3: 2

   
   
   
   

 

Source

The 36th ACM/ICPC Asia Regional Shanghai Site —— Warmup

 

广搜题目，枚举起点，染色为对的颜色，然后向左至对的颜色，向右至对的颜色  然后上下两行一起

状态压缩   每行最多八个点  所以2^16可以将状态存储

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

int t,n,tt;
bool vis[2][8],tmp[2][8];
char ch[2][8];
bool flag[1<<16];//标记判重
struct node
{
    int flag;
    int t;
} st,ed;

void set_vis(node a)  //获得状态矩阵值
{
    memset(vis,false ,sizeof(vis));
    int i=0;
    while(i<n)
    {
        vis[0][i]=(a.flag&1);
        a.flag/=2;
        i++;
    }
    i=0;
    while(i<n)
    {
        vis[1][i]=(a.flag&1);
        a.flag/=2;
        i++;
    }
    return ;
}

int get_flag()  //获得状态压缩值
{
    int ans=0;
    for(int i=0; i<n; i++)
    {
        if(tmp[0][i])
            ans|=(1<<i);
        if(tmp[1][i])
            ans|=(1<<(i+n));
    }
    return ans;
}

int l,r;
void set_tmp(int x,int y)  //找到最大改变区间
{
    l=y,r=y;
    while(l>=0&&!vis[x][l])
        l--;
    while(r<n&&!vis[x][r])
        r++;
    for(int i=l+1; i<r; i++)
        tmp[x][i]=(ch[x][y]==ch[x][i]);
    return ;
}

void bfs()
{
    queue<node>q;
    q.push(st);
    while(!q.empty())
    {
        st=q.front();
        q.pop();
        set_vis(st);
        if(st.flag==(1<<(n*2))-1)  //全部染色对的情况
        {
            printf("Case #%d: %d\n",tt++,st.t);
            return ;
        }
        for(int i=0; i<2; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(!vis[i][j])
                {
                    memcpy(tmp,vis,sizeof(vis));
                    set_tmp(i,j);
                    //枚举染色点
                    for(int k=l+1;k<r;k++)
                    {
                        tmp[i][k]=(ch[i][k]==ch[i][j]);
                        ed.flag=get_flag();
                        if(!flag[ed.flag])
                        {
                            ed.t=st.t+1;
                            flag[ed.flag]=true ;
                            q.push(ed);
                        }
                    }
                    memcpy(tmp,vis,sizeof(vis));
                    //两行
                    for(int k=l+1;k<r;k++)
                    {
                        tmp[i][k]=(ch[i][k]==ch[i][j]);
                        tmp[i^1][k]=(ch[i^1][k]==ch[i][j]);
                        ed.flag=get_flag();
                        if(!flag[ed.flag])
                        {
                            ed.t=st.t+1;
                            flag[ed.flag]=true ;
                            q.push(ed);
                        }
                    }
                }
            }
        }
    }
    return ;
}

int main()
{
    cin>>t;
    tt=1;
    while(t--)
    {
        cin>>n;
        cin>>ch[0]>>ch[1];
        memset(flag,0,sizeof(flag));
        st.flag=0;
        st.t=0;
        flag[0]=true ;
        bfs();
    }
    return 0;
}