ACWING 282-Stone merge (interval DP)

Address: https://www.acwing.com/problem/content/284/

 

 

 　　　　Topic: Give a pile of stones, adjacent mergers, ask the minimum cost, the specific algorithm is in the question.

　　　  Resolution:

　　　　　　  For this question, according to common sense, for the final state, the two groups merged. So define dp [i] [j] to represent the minimum value of the interval ij merger.

　　　　　　  Two piles merge: dp [1] [2] = dp [1] [1] + dp [2] [2] + sum [1 ~ 2];

　　　　　　　　　       即dp[i][i+1]=dp[i][i]+dp[i+1][i+1]+sum[i~i+1];

　　　　        Three piles combined: dp [1] [3] = min (dp [1] [1] + dp [2] [3], dp [1] [2] + dp [3] [3]) + sum [1 ~ 3];

　　　　　　　　　     dp[i][i+2]=min(dp[i][i]+dp[i+1][i+2],dp[i][i+1]+dp[i+2][i+2])+sum[i~i+2];

　　　　       In summary, it is extended to the merge of the i-th pile to the j-th pile, dp [i] [j] = min (dp [i] [j], dp [i] [k] + dp [k + 1] [j] + sum [j] -sum [i-1]);

　　　　       Three fors is equivalent to traversing each length interval.

　　　　       It feels a bit floyed ... Find a transit point

#include<iostream>
#include<cstdio>
#include<cstring>
const int maxn = 500;
const int inf = 1e9;
int dp[maxn][maxn];
int a[maxn],sum[maxn];
using namespace std;
int main()
{
    int n;
    cin>>n;
    sum[0]=0;
    for(int i = 1;i <= n ; i++)
    {
        cin>>a[i];
        sum[i]=sum[i-1]+a[i];
    }
    for(int len =2 ; len <= n ;len++)
    {
        for(int l = 1; l+len-1<=n; l++)
        {
            int r=l+len-1;
            dp[l][r]=inf;
            for(int k = l ; k  < r ; k ++)
            {
                dp[l][r]=min(dp[l][r],dp[l][k]+dp[k+1][r]+sum[r]-sum[l-1]);
            }
        }
    }
    cout<<dp[1][n]<<endl;
}