二维状态压缩——cf903F

Posted 2021-03-10 zsben991126

题解看这个https://blog.csdn.net/u013534123/article/details/78926568

状压的debug真的很烦

#include<bits/stdc++.h>
using namespace std;

int mask[5][5],n,a[5],dp[1010][1<<16],b[5][2005];
char mp[5][2005];
/*
15 11 7 3
14 10 6 2
13 9  5 1
12 8  4 0
*/
void premask(){
    mask[0][0]=0b1111111111111111;
    mask[0][1]=0b0111111111111111;
    
    mask[1][0]=0b1111111111111111;
    mask[1][1]=0b1011111111111111;
    mask[1][2]=0b0011001111111111;
    
    mask[2][0]=0b1111111111111111;
    mask[2][1]=0b1101111111111111;
    mask[2][2]=0b1001100111111111;
    mask[2][3]=0b0001000100011111;
    
    mask[3][0]=0b1111111111111111;
    mask[3][1]=0b1110111111111111;
    mask[3][2]=0b1100110011111111;
    mask[3][3]=0b1000100010001111;
    mask[3][4]=0b0000000000000000;
}

int main(){
    cin>>n;
    for(int i=1;i<=4;i++)cin>>a[i];
    for(int i=0;i<4;i++)scanf("%s",mp[i]);
    for(int i=0;i<4;i++)
        for(int j=0;j<n;j++)
            b[i][j]=mp[i][j]==‘*‘?1:0;
    
    memset(dp,0x3f,sizeof dp);

    
    int init=0;
    for(int j=0;j<4;j++)
        for(int i=0;i<4;i++)
            init<<=1,init|=b[i][j];
    dp[0][init]=0;
    
    premask();
    
    for(int i=0;i<n;i++){
        for(int s=0;s<(1<<16);s++){
            if(dp[i][s]==0x3f3f3f3f)continue;
            for(int l1=0;l1<=1;l1++)
                for(int l2=0;l2<=2;l2++)
                    for(int l3=0;l3<=3;l3++)
                        for(int l4=0;l4<=4;l4++){
                            int now=s;
                            //cout<<i<<" "<<now<<‘
‘;
                            now&=mask[0][l1];
                            //cout<<i<<" "<<now<<‘
‘;
                            now&=mask[1][l2];
                            //cout<<i<<" "<<now<<‘
‘;
                            now&=mask[2][l3];
                            //cout<<i<<" "<<now<<‘
‘;
                            now&=mask[3][l4];
                            //cout<<i<<" "<<now<<‘
‘;
                            //cout<<"
";
                            if((now>>15)&1)continue;
                            if((now>>14)&1)continue;
                            if((now>>13)&1)continue;
                            if((now>>12)&1)continue;
                            
                            int cost=a[l1]+a[l2]+a[l3]+a[l4];
                            int nxt=now;
                            for(int j=0;j<4;j++){
                                nxt<<=1;
                                nxt&=(1<<16)-1;
                                nxt|=b[j][i+4];
                            }
                            dp[i+1][nxt]=min(dp[i+1][nxt],dp[i][s]+cost);
                            /*cout<<l1<<" "<<l2<<" "<<l3<<" "<<l4<<‘
‘;
                            cout<<i<<" "<<s<<" "<<nxt<<"
";
                            cout<<dp[i][s]<<" "<<dp[i+1][nxt]<<"
";
                            */
                        }
        }
    }
    /*
    for(int i=0;i<n;i++){
        for(int s=0;s<(1<<16);s++)
            if(dp[i][s]!=0x3f3f3f3f)cout<<dp[i][s]<<" ";
        puts("");
    }*/
    cout<<dp[n][0]<<" ";
}