Codeforces1071c triple flips [constructor] [four Russians]

Question Analysis:

Which of the following questions can be considered first. We found that we can choose three consecutive operations without considering the number of operations.

In this way, we can always convert a sequence into the form of $000... 000xy $. In other words, for $000... 0001 $, $000... 0010 $, $000... 0011 $, consider no solution conditions.

At this time, write a brute-force program and use the $ O (N ^ 2*2 ^ N) $ algorithm to find that the total length is less than $7 $, the other two values are less than $8 $.

Then, observe that the number of operations required by the question is related to $ N/3 $. It is not difficult to think of every $3 $ into a group, and then change it to $0 $ in the group to allow the use of out-of-group information, each group can only be operated once. In this case, the number of operations exceeds ($011 or 110 $ ).

Then, let's relax the limit. Every $6 $ is divided into one group and adjusted to $0 $ in the group. You can use the information outside the group. Each group can only be operated twice. In this way, I thought that $110111 was not acceptable at first. After I wrote a brute force code, I found my brain dead ).

In this case, we will consider four Russians, pre-process how to proceed to each $6 $ group, and finally we will be able to defend against the remaining violence.

 

Code:

  1 #include<bits/stdc++.h>  2 using namespace std;  3   4 const int maxn = 102000;  5   6 int n,a[maxn],f[1<<12];  7 struct node{int x,y,z;};  8 vector <node> des[64];  9  10 void init(){ 11     queue<int> q; 12     int N = 6; 13     memset(f,0x3f,sizeof(f)); 14     f[0] = 0; q.push(0); 15     while(!q.empty()){ 16     int k = q.front();q.pop(); 17     for(int i=0;i<N;i++){ 18         if(f[k^(1<<i)] > f[k] + 1){ 19         for(int j=0;j<des[k].size();j++) 20             des[k^(1<<i)].push_back(des[k][j]); 21         des[k^(1<<i)].push_back((node){i,6,12-i}); 22         f[k^(1<<i)] = f[k]+1; 23         q.push(k^(1<<i)); 24         } 25         for(int j=i+1;j<N;j++){ 26         int z = (1<<i)+(1<<j); 27         if(j+(j-i) < N) z |= (1<<2*j-i); 28         if(f[k^z] > f[k]+1){ 29             for(int fr=0;fr<des[k].size();fr++) 30             des[k^z].push_back(des[k][fr]); 31             des[k^z].push_back((node){i,j,2*j-i}); 32             f[k^z] = f[k]+1; 33             q.push(k^z); 34         } 35         } 36     } 37     } 38 } 39  40 node opt[maxn]; 41 int num; 42  43 node lst[maxn]; 44  45 void Remain(){ 46     queue<int> q; 47     memset(f,0x3f,sizeof(f)); 48     //for(int i=0;i<64;i++) des[i].clear(); 49     f[0] = 0;q.push(0); 50     int N = min(n,12); 51     while(!q.empty()){ 52     int k = q.front();q.pop(); 53     for(int i=0;i<N;i++){ 54         for(int j=i+1;j<N;j++){ 55         if(j+(j-i) >= N) break; 56         int z = (1<<i)+(1<<j)+(1<<2*j-i); 57         if(f[k^z] > f[k]+1){ 58             f[k^z] = f[k]+1; 59             lst[k^z] = (node){i,j,2*j-i}; 60             q.push(k^z); 61         } 62         } 63     } 64     } 65     int z = 0; 66     for(int i=n-N+1;i<=n;i++){z = (z<<1)+a[i];} 67     if(f[z] > 1e8){puts("NO");} 68     else{ 69     puts("YES"); 70     printf("%d\n",num+f[z]); 71     for(int i=1;i<=num;i++){ 72         printf("%d %d %d\n",opt[i].x,opt[i].y,opt[i].z); 73     } 74     while(z){ 75         printf("%d %d %d\n",n-lst[z].z,n-lst[z].y,n-lst[z].x); 76         int pp = z; 77         z ^= (1<<lst[pp].x); z ^= (1<<lst[pp].y); z^= (1<<lst[pp].z); 78     } 79     } 80 } 81  82 void work(){ 83     for(int i=1;i<=n;i+=6){ 84     if(i+12 > n) break; 85     int z = (a[i+5]<<5)+(a[i+4]<<4)+(a[i+3]<<3)+(a[i+2]<<2)+(a[i+1]<<1)+a[i]; 86     for(int j=0;j<des[z].size();j++){ 87         opt[++num] = des[z][j]; 88         opt[num].x += i; opt[num].y += i; opt[num].z += i; 89         a[opt[num].x] ^= 1; a[opt[num].y] ^= 1; a[opt[num].z] ^= 1; 90     } 91     } 92     Remain(); 93 } 94  95 int main(){ 96     scanf("%d",&n); 97     for(int i=1;i<=n;i++) scanf("%d",&a[i]); 98     init(); 99     work();100     return 0;101 }

 

Codeforces1071c triple flips [constructor] [four Russians]