1068. Consolidation of ring stones (ring, interval DP)

Original link - AcWing 

analysis ： Merging with stones only adds a circular condition

For the ring problem , If you just enumerate breakpoints , Open more one dimension on the basis of stone merging , The time complexity is O(n^4)

A general method can be adopted , Is to convert a ring into a chain

 

  By turning from 1 The expanded link is in n Behind , When enumerating like this, Enumerate with Location 2 Broken chain , Just enumerate With 2 Start , The length is len Chain , You can enumerate n And 1 Adjacent situation

The state transition equation is basically the same as the stone merging , The difference lies in the existence of rings , Cause the loop enumeration to be different

for(int len=1;len<=n;len++)
        for(int l=1;l+len-1<=2*n;l++)
        {// Due to the simulation of intermediate breakpoints , Connect the disconnected ones to the back of the array , So it is possible to enumerate to 2n

           ...
        }

 

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=410;
int a[N];
int s[N];
int dp[N][N];
int g[N][N];

int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i],a[n+i]=a[i];
    
    for(int i=1;i<=2*n;i++) s[i]=s[i-1]+a[i];
    
    memset(dp,0x3f,sizeof dp),memset(g,-0x3f,sizeof g);
    
    for(int len=1;len<=n;len++)
        for(int l=1;l+len-1<=2*n;l++)
        {
            int r=l+len-1;
            if(len==1) dp[l][r]=g[l][r]=0;
            else
                for(int k=l;k<r;k++)
                {
                    dp[l][r]=min(dp[l][r],dp[l][k]+dp[k+1][r]+s[r]-s[l-1]);
                    g[l][r]=max(g[l][r],g[l][k]+g[k+1][r]+s[r]-s[l-1]);
                }
        }
    int maxv=-0x3f3f3f3f,minv=0x3f3f3f3f;
    for(int i=1;i<=n;i++)
        minv=min(minv,dp[i][i+n-1]),maxv=max(maxv,g[i][i+n-1]);
    cout<<minv<<endl<<maxv<<endl;
    return 0;
    
}