HDU 2844 Coins Multiple backpack

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Problem Solving Ideas:

Multi-knapsack problem, not very understanding, understand the supplementary.

The idea of an online blog is:

for each coin: IF value x number >=m        then the number of times to take this coin is equal to unlimited, can be considered as a complete backpack ELSE then        consider 0-1 backpack (binary optimization), is to combine the V and num of this coin outof the possible state of the 0-1 backpack (which can be seen in the Backpack IX) (for NUM, which is similar to encoding.)  when 2 2^n (n=0,1,2, ...) Indicates a state, which corresponds to a number of binary numbers is 1, and then there is a state is K>NUM/2 time, num+1-K, so down can use K to assemble all possible from the 1->num. Then for K, the unit value and size are multiplied by K and become a 0-1 backpack)

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The source code is as follows:

#include <stdio.h>#include<algorithm>#include<string.h>using namespacestd; Const intmax=100000; intDp[max]; intC[max],w[max]; intv; voidZeroonepack (intCostintWei// on{      inti;  for(i = v;i>=cost;i--) {Dp[i]= Max (dp[i],dp[i-cost]+Wei); }  }    voidCompletepack (intCostintWei//completely{      inti;  for(i = cost;i<=v;i++) {Dp[i]= Max (dp[i],dp[i-cost]+Wei); }  }    voidMultiplepack (intCostintWeiintCnt//multiple{      if(v<=cnt*Cost )          {Completepack (Cost,wei); return ; }      Else    {          intK =1;  while(k<=CNT) {Zeroonepack (k*cost,k*Wei); CNT= cnt-K; K=2*K; } zeroonepack (CNT*cost,cnt*Wei); }  }    intMain () {intN;  while(~SCANF ("%d%d", &n,&v), n+v) {inti;  for(i =0; i<n;i++) scanf ("%d",&C[i]);  for(i =0; i<n;i++) scanf ("%d",&W[i]); Memset (DP,0,sizeof(DP));  for(i =0; i<n;i++) {multiplepack (c[i],c[i],w[i]); }          intsum =0;  for(i =1; i<=v;i++)          {              if(dp[i]==i) {sum++; }} printf ("%d\n", sum); }      return 0; }  

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HDU 2844 Coins Multiple backpack