Luo Gu P1043 (water problem solution to a problem)

First, beginning with the first AC codes (DP Gangster Act of original links )

#include<bits/stdc++.h>
using namespace std;
int n,m,f1[110][110][110],f2[110][110][110],a[110],sum[110];
int mod(int x){
    return (x%10+10)%10;
}
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){cin>>a[i];sum[i]=sum[i-1]+A [I];}
     for ( int I = . 1 ; I <= n-; I ++) {A [I + n-] = A [I], SUM [I + n-] = SUM [I] + SUM [n-]; } 
    Memset (F2, 127 , the sizeof (F2)); // F2 record is initialized to a minimum value so that a maximum value 
    for ( int I = . 1 ; I <= 2 * n-; I ++ ) {
         for ( int J = I; J <= 2 * n-; J ++ ) { 
            [I] F1 [J] [ . 1 ] F2 = [I] [J] [ . 1 ] = MOD (SUM [J] -sum [I- . 1 ]); / / pretreatment each segment value 
        } 
    } 
    for ( intL = . 1 ; L <= n-; L ++) { // interval length 
        for ( int I = . 1 , J = I + L- . 1 ; J <= 2 * n-; I ++, J ++) { // left endpoint 
            for ( int H = 2 ; H <= m; H ++) { // the number of segments 
                for ( int K = I + H- . 1 ; K <J; K ++) { // short point 
                    f1 [i] [j] [ h] = max ( F1 [I] [J] [H], F1 [I] [K] [H- . 1 ] * MOD (SUM [J] - SUM [K])); 
                    F2 [I] [J] [H] = min (F2 [I] [J], F2 [I] [H] [K] [H- . 1 ] * MOD (SUM [J] -  SUM [K]));
                }
            }
        }
    }
    int maxn=0,minn=0x7fffffff;
    for(int i=1;i<=n;i++){//找答案啦～～～
        maxn=max(maxn,f1[i][i+n-1][m]);
        minn=min(minn,f2[i][i+n-1][m]);
    }
    cout<<minn<<endl<<maxn;
}

Gangster ideas

Set state

DP routine is entitled to ask what is what is located, then there is a range of DP this classic title

So we let F [i] [j] [h] h represent the maximum and minimum divided segments from i to j

Transfer equation

We can draw from the state transition equation:

Enumeration first interval length

DP from the normal range of templates we know

why:

If we first enumerate i, j

For example, i = 1, j = 10, k = 5 when

f [1] [5] [h] it has not been so processed out wrong

And then enumerate about endpoint

This is easy to understand, interval DP Well

Re-enumeration is the number of segments

This is also a good thing to understand the answer to be updated