Search --UCF2015 H

Look like a title feel like artillery lineup similar pressure

Write for a long time found to be false solution: This problem has a stronger restriction is to fill,

　Therefore, like the case to use pressure dp dp [i] [s] [ss] represents the state of color and fills policy i-th row,

　Or the outer layer 3 ^ n sub enumerated set of complexity do

    This way figured not to the time complexity O (3 ^ n * 2 ^ n * n)

In fact, you can use direct search dfs pruning

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

char Map[10][10];
int T,N,M,Ans,Case=0;

void DFS(int i,int j,int Step,int Count){
    if(Step>=Ans) return;
    if(Count==0){Ans=min(Ans,Step);return;}
    if(i==1 || i>=N || j==1 || j>=M) return;
    if((Count-1)/5+1+Step>=Ans) return;
    char temp[5];
    int Cnt=0;bool flag=false;
    temp[0]=Map[i][j];
    temp[1]=Map[i-1][j];temp[2]=Map[i+1][j];
    temp[3]=Map[i][j-1];temp[4]=Map[i][j+1];
    for(register int k=0;k<5;++k){
        if(temp[k]=='#') ++Cnt;
        if(temp[k]=='.'){flag=true;break;}
    }
    if(!flag && Cnt){
        Map[i][j]=Map[i-1][j]=Map[i+1][j]=Map[i][j-1]=Map[i][j+1]='@';
        if(j+1<M) DFS(i,j+1,Step+1,Count-Cnt);
        else DFS(i+1,2,Step+1,Count-Cnt);
        Map[i][j]=temp[0];
        Map[i-1][j]=temp[1];Map[i+1][j]=temp[2];
        Map[i][j-1]=temp[3];Map[i][j+1]=temp[4];
    }
    if(i==2 && Map[1][j]=='#') return;
    if(j==2 && Map[i][1]=='#') return;
    if(i==N-1 && Map[N][j]=='#') return;
    if(j==M-1 && Map[i][M]=='#') return;
    if(j+1<M) DFS(i,j+1,Step,Count);
    else DFS(i+1,2,Step,Count);
    return;
}

int main () {
    ios::sync_with_stdio(false);
    cin>>T;
    while(T--){
        ++Case;
        cout<<"Image #"<<Case<<": ";
        cin>>N>>M;
        Ans=2147483647;
        int Count=0;
        for(register int i=1;i<=N;++i){
            for(register int j=1;j<=M;++j){
                cin>>Map[i][j];
                if(Map[i][j]=='#') ++Count;
            }
        }
        if(Map[1][1]=='#'||Map[1][M]=='#'||Map[N][1]=='#'||Map[N][M]=='#'){
            cout<<"impossible"<<endl<<endl;
            continue;
        }
        DFS(2,2,0,Count);
        if(Ans==2147483647) cout<<"impossible"<<endl;
        else cout<<Ans<<endl;
        cout<<endl;
    }
    
    return 0;
}