Topic description
Place N piles of stones around a circular playground, and now combine the stones into a pile in an orderly manner. It is stipulated that only 2 adjacent piles can be selected to merge into a new pile at a time, and the new pile of stones can be merged into a new pile. number, recorded as the combined score.
Try to design an algorithm to calculate the minimum and maximum scores for combining N piles of stones into one pile.
Input and output format
Input format:The first row of the data is a positive integer N, 1≤N≤100, which means there are N piles of stones. The second row has N numbers, which respectively represent the number of stones in each pile.
Output format:The output has a total of 2 lines, the first line has the minimum score, and the second line has the maximum score.
Input and output example
Solution：
This question can not be greedy, you can find it casually.
Considering the interval $DP$, let $f[i][j]$ represent the cost of combining $ij$, and $sum[i]$ represent the prefix sum of costs to $i$. Since it is a ring, it is directly broken into a chain in the original Copy the original array after the array, and it is not difficult to get the state transition equation: $f[i][j]=max(f[i][j],f[i][k]+f[k+1][j ]+s[j]-s[i-1]),k\in[i,j)$, pay attention to the boundary $f[i][i]=0,f[i][i+1]=s[ i+1]-s[i-1]$, recursive enumeration interval length, because it is a ring, so the target state is the minimum\maximum value in $f[i][n+i-1]$.
Code:
1 #include<bits/stdc++.h> 2 #define il inline 3 #define ll long long 4 #define Min(a,b) (a)>(b)?(b):(a) 5 #define Max(a,b) (a)>(b)?(a):(b) 6 #define f_for(a,b,i) for(int (i)=(a);(i)<=(b);(i)++) 7 using namespace std; 8 const int N=205; 9 int n,a[N],s[N],f1[N][N],f2[N][N],ans1=520520520,ans2; 10 int main(){ 11 ios::sync_with_stdio(0); 12 cin>>n; 13 memset(f1,0x3f,sizeof(f1));memset(f2,-0x3f,sizeof(f2)); 14 f_for(1,n,i)cin>>a[i],a[i+n]=a[i],f1[i][i]=f2[i][i]=f1[i+n][i+n]=f2[i+n][i+n]=0; 15 f_for(1,(n<<1)-1,i)s[i]=a[i]+s[i-1]; 16 f_for(1,n-1,l) 17 f_for(1,(n<<1)-1,i){ 18 int j=i+l; 19 if(j>((n<<1)))break; 20 if(l==1)f1[i][j]=f2[i][j]=s[j]-s[i-1]; 21 else f_for(i,j-1,k) 22 f1[i][j]=Min(f1[i][j],f1[i][k]+f1[k+1][j]+s[j]-s[i-1]), 23 f2[i][j]=Max(f2[i][j],f2[i][k]+f2[k+1][j]+s[j]-s[i-1]); 24 } 25 f_for(1,n,i) 26 if((i+n-1)>((n<<1)))break; 27 else ans1=Min(ans1,f1[i][i+n-1]),ans2=Max(ans2,f2[i][i+n-1]); 28 cout<<ans1<<endl<<ans2; 29 return 0; 30 31 }