Beef computer memory (like pressure DP)

Links: https://ac.nowcoder.com/acm/problem/21873

Source: Cattle-off network

Title Description

Taurus a total of m blocks of computer memory, there are n instructions, each instruction is a sequence of 01
if the i-th character is a command, indicating that instruction requires access to memory block i
execution cost of each instruction is k ^ 2, k is the number of new access memory, that is before the instruction does not have access to the amount of memory
you can freely arrange the order of execution of the instruction n, find complete instructions to perform all of the minimum cost

Enter a description:

The first input line of a first integer n-( $\mathrm{.\; 1\; \le \; n-\; \le \; 20\; is\; 1\le n\le 20)\; Next\; per\; line\; 01\; a\; string\; of\; not\; more\; than\; 20}$

Output Description:

An integer output

Specific ideas:

DP [i] denotes the i which corresponds to the selected binary lines of minimum cost.

Val [i] denotes the i which corresponds to the selected binary lines corresponding to the memory access which

AC Code:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 # define ll long long
 4 # define inf 0x3f3f3f3f
 5 const int mod = 1e9+7;
 6 const int maxn = 2e6+100;
 7 int dp[maxn],val[maxn];
 8 int n,m;
 9 int cal(char str[])
10 {
11     int len=strlen(str);
12     int tmp=1;
13     int ans=0;
14     for(int i=len-1; i>=0; i--)
15     {
16         ans+=(str[i]-'0')*tmp;
17         tmp<<=1;
18     }
19     return ans;
20 }
21 char str[25];
22 int f(int t1,int t2)
23 {
24     int ans=0;
25     for(int i=0; i<=20; i++)
26     {
27         if(((t1&(1<<i))==0)&&(t2&(1<<i)))
28             ans++;
29     }
30     return ans*ans;
31 }
32 int a[maxn];
33 int main()
34 {
35     memset(dp,inf,sizeof(dp));
36     dp[0] = 0 ;
 37 [      Scanf ( " % D% D " , & n-, & m);
 38 is      for ( int I = 0 ; I <n-; I ++ )
 39      {
 40          Scanf ( " % S " , STR);
 41 is          A [ I] = CAL (STR); // because when comparing the comparison is a decimal, the first converted into binary
 42 is      }
 43 is      int maxstate = ( . 1 << n-) - . 1 ;
 44 is      for ( int I = 0 ; I < maxstate =; I ++ )
 45      {
46         for(int j=0; j<n; j++)
47         {
48             if(i&(1<<j))
49                 continue;
50             if(dp[i]==inf)
51                 continue;
52             int tmp=dp[i]+f(val[i],a[j]);
53             if(dp[i|(1<<j)]>tmp)
54             {
55                 dp[i|(1<<j)]=tmp;
56                 val[i|(1<<j)]=val[i]|a[j];
57             }
58         }
59     }
60     printf("%d\n",dp[maxstate]);
61     return 0;
62 }