[MSSB]缩小枚举范围后的简易递推

特征

• 每一个位置顶多只会操作一次。如果操作两次的话，相当于不操作，必然不满足最优解

• 在一套方案中，操作的顺序无关紧要

• 如果我们确定了第1行的操作方案的话，那么后面的行数都可以依此递推

解法

枚举第一行的操作方案，然后依次递推
时间复杂度为\(O(1<<len(1))\)

题目

费解的开关

#include <bits/stdc++.h>
using namespace std;
int n,m,i,j,k,a[7][7],ans1=1e6,b[7][7];
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        getchar();
        for (i=1;i<=5;i++)
        {
            for (j=1;j<=5;j++)
            {
                char ch=getchar();
                b[i][j]=ch-'0';
            }
            getchar();
        }
        for (i=0;i<=(1<<5);i++)
        {
            for (j=1;j<=5;j++)
            {
                for (k=1;k<=5;k++)
                    a[j][k]=b[j][k];
            }
            int ans=0;
            for (j=1;j<=5;j++)
                if (i>>(j-1) & 1)
                {
                    ans++;
                    a[1][j-1]^=1;
                    a[1][j+1]^=1;
                    a[1][j]^=1;
                    a[2][j]^=1;
                }
            for (j=1;j<=4;j++)
                for (k=5;k>=1;k--)
                    if (!a[j][k])
                    {
                        ans++;
                        a[j][k]^=1;
                        a[j+2][k]^=1;
                        a[j+1][k]^=1;
                        a[j+1][k+1]^=1;
                        a[j+1][k-1]^=1;
                    }
            bool ok=true;
            for (j=1;j<=5;j++)
                for (k=1;k<=5;k++)
                    if (!a[j][k])
                        ok=false;
            if (ok)
                ans1=min(ans1,ans);
        }
        if (ans1>6)
            cout<<-1;
        else
            cout<<ans1;
        ans1=1e7;
        puts("");
    }
    return 0;
}

翻转游戏

#include<bits/stdc++.h>
using namespace std;
#define loop(i,start,end) for(register int i=start;i<=end;++i)
#define anti_loop(i,start,end) for(register int i=start;i>=end;--i)
#define clean(arry,num) memset(arry,num,sizeof(arry))
#define ll long long
#define pd_one(pos,num) (((1<<(pos-1))&num)!=0)
#define copy(arry,arry2) memcpy(arry,arry2,sizeof(arry2)) 
template<typename T>void read(T &x){
	x=0;char r=getchar();T neg=1;
	while(r>'9'||r<'0'){if(r=='-')neg=-1;r=getchar();}
	while(r>='0'&&r<='9'){x=(x<<1)+(x<<3)+r-'0';r=getchar();}
	x*=neg;
}


int n;
const int maxn=20;
char MAP[maxn][maxn];
int G[maxn][maxn];
int op[maxn][maxn];

inline void reset(int x,int y){
	op[x][y]^=1;
	op[x-1][y]^=1;
	op[x+1][y]^=1;
	op[x][y+1]^=1;
	op[x][y-1]^=1;
}

inline void printop(){
	printf(">>>>>>>>>>>>>>>>>>>>>>>>printop\n");
	loop(i,1,n){
		loop(j,1,n)
			printf("%d ",op[i][j]);
		printf("\n");
	}
}

inline bool judge(int a){
	loop(i,1,n)
		if(op[n][i]==a){
			return false;
		}
	return true;
}

int main(){
	#ifndef ONLINE_JUDGE
	freopen("datain.txt","r",stdin);
	#endif
	
	read(n);
	loop(i,0,n-1)
		scanf("%s",MAP[i]);
	loop(i,1,n){
		loop(j,1,n){
			G[i][j]=(MAP[i-1][j-1]=='b')?1:0;
		}
	}
	
	int result=INT_MAX;
	for(int state=0;state<(1<<n);++state){
		int res=0;
		copy(op,G);
		//printop();
		loop(i,1,n){
			if(pd_one(i,state)){
				reset(1,i);
				++res;
				//printop();
			}
		}
		
		//printop();
		//printf("res:%d\n",res);
		
		loop(i,1,n-1){//hang
			loop(j,1,n){//lie
				if(op[i][j]==0){
					reset(i+1,j);
					++res;
					//printop();
				}
			}
		}
		
		if(judge(0))
			result=min(res,result);
		//>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
		res=0;
		copy(op,G);
		
		loop(i,1,n){
			if(pd_one(i,state)){
				reset(1,i);
				++res;
				//printop();
			}
		}
		
		loop(i,1,n-1){//hang
			loop(j,1,n){//lie
				if(op[i][j]==1){
					reset(i+1,j);
					++res;
					//printop();
				}
			}
		}
		
		loop(i,1,n)
			if(op[n][i]){
				res=INT_MAX;
				break;
			}
		
		if(judge(1))
			result=min(res,result);
	}
	if(result<INT_MAX)
		printf("%d\n",result);
	else printf("Impossible\n");
	return 0;
}