There backpack Los dependent Valley P1064 Jinming budget plan (not grouped backpack)

 

Meaning of the questions: buy accessories premise is to buy the main pieces, and asked to get the most value for a given amount.

 

Analysis: dependent knapsack problem template, note that there are not dependent backpack backpack grouping of three for the kind of order, but a post-enumeration of the main pieces, constantly enumeration accessory, purchase a number of different accessories.

 

Meanwhile, in writing, with the f array represents an array of answers, each j f is the corresponding optimal decision-making, and each enumerate all operations on the array t, and then determine what needs to update the f array.

  * Packet backpack is an article grouping, each group of items conflict with each other, can only choose one item in a bag. In fact, the "select all items in a" from become a "select group from the current one", so they each group to conduct a knapsack on it.

  * Packet backpack core code:

for ( int K = . 1 ; K <= TS; K ++)           // cycle each group 
  for ( int I = m; I> = 0 ; i--)          // cycle backpack capacity 
    for ( int J = . 1 ; J < = CNT [K]; J ++)   // for each article cycle of the group 
      IF (I> = W [T [K] [J]]) 
        DP [I] = max (DP [I], 
                    DP [I - W [T [K] [J]]] + C [T [K] [J]]);   // as knapsack as the state transition, the same as the 0-1 knapsack prevent duplicate into the same article ideas

 

This problem Code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;

const int maxn = 202003;
struct node{
    int p,v,f;
}a[777];
int f[maxn];
int t[maxn];

int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1; i<=m; i++){
        scanf("%d%d%d",&a[i].p,&a[i].v,&a[i].f);
        a[i].v *= a[i].p;
    }
    for(int i=1; i<=m; i++){               //一步
        if(!a[i].f){
            for(int j=0; j<a[i].p; j++){
                t[j] = f[j];
            }
            for(int j=a[i].p; j<=n; j++){
                t[j] = f[j - a[i].p]+a[i].v;
            }
            for( Int K = . 1 ; K <= m; K ++) {       // two steps 
                IF (A [K] .F == I) {
                     for ( int J = n-; J> = A [I] .p; J- -) {     // three steps (note that this is not written a plurality of sets of the backpack, the article can be put into the same group)   
                        IF (JA [K] .p> = a [I] .p) {
                             // the printf ( "Yes \ n-"); 
                            T [J] = max (T [JA [K] .p] + A [K] .v, T [J]);
                             // the printf ("% D% D \ n-", J, T [J]); 
                        } 
                    } 
                } 
            } 
            for ( int j=a[i].p; j<=n; j++){
                f[j] = max(f[j],t[j]);
            }
        }
    }
    printf("%d\n", f[n]);

}