Dynamically plan the appearance level (mg) Questions and solutions, and mg questions

Dynamically plan the appearance level (mg) Questions and solutions, and mg questions

A company carries a batch of boxes in N (1 ≤ N ≤ 1000) in sequence by "conveyor belt", and then arranges at most P (1 ≤ P ≤ 4) columns in the warehouse, (8 ). Currently, we know that a maximum of M (1 ≤ M ≤ 20) boxes can be shipped. We want to sort the boxes of the same type together as much as possible to make them beautiful. "Beauty degree" T is defined as: T = Σ (the number of different types seen in each column in sequence); the so-called "Number of different types seen in sequence" is: if the K box in a column is different from the K-1 box type, the value of "Beauty level" is increased by 1. Find a transfer arrangement to minimize the "Beauty level" value of each column .! Http://img.blog.csdn.net/20150828211629851)

The difficulty of this question is how to compress the status. We can use m + 1 hexadecimal numbers for State compression. Then, a dfs function is used to search for the status for status transfer. F [I] [j] Indicates the minimum aesthetic level when the status of each column of the First I number is compressed to j.
Additional code (original code by TA god. Reprinted with the source ):

#include<cstdio>#include<cstring>#include<iostream>using namespace std;#include<algorithm>#define S 21int f[2][194485],p,now,pre=1,mi[5],m,a;void out(int state){    for(int i=0;i<4;++i)cout<<state%mi[i+1]/mi[i]<<" ";    cout<<endl;}void dfs(int x,int state){    if(x<0){        if(f[pre][state]<1000){            int i;            for(i=p;i--;){                x=state%mi[i+1]/mi[i];                f[now][state+(a-x)*mi[i]]=min(f[now][state+(a-x)*mi[i]],f[pre][state]+(x!=a));            }        }        return;    }    for(int i=m;~i;--i)dfs(x-1,state*(m+1)+i);}int main(){    freopen("mg.in","r",stdin);    freopen("mg.out","w",stdout);    int n,i;    scanf("%d%d%d",&n,&m,&p);    mi[0]=1;    for(i=1;i<5;++i)mi[i]=mi[i-1]*(m+1);    memset(f,127,sizeof(f));    f[0][0]=0;    while(n--){        scanf("%d",&a);        swap(now,pre);        memset(f[now],127,sizeof(f[now]));        dfs(p,0);    }    printf("%d\n",*min_element(f[now],f[now+1]));}

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