HDU 2191 Mourning 512 Wenchuan earthquake--cherish now, Thanksgiving life (multiple backpack)

Test instructions: Give the maximum value of n, and then give the type M and each bag number C, price p, weight h, ask to buy the maximum weight of rice

Idea: Each item has a fixed maximum number of times. For multiple knapsack problems. Convert to 01 backpack to do

The following methods are converted to 01 backpack to do

Idea 1: The item does not spread out, choose each kind of time, carries on the discussion, (equivalent to the vertical filling backpack v The column, one column one column fills

Code 1:KJ

#include <iostream>#include<stdio.h>#include<string.h>using namespacestd;intdp[ the];intMain () {intI,j,k,tem; intT,n,_v;//number of test cases, item type, backpack size    intc[ the],v[ the];//Cost, Value    intnum[ the];//number of items per item    intTC,TV;//cost of Split items, valuescanf"%d",&t);  while(t--) {scanf ("%d%d",&_v,&N); Memset (DP,0,sizeof(DP));  for(i=1; i<=n; ++i) scanf ("%d%d%d",&c[i],&v[i],&Num[i]); //////         for(i=1; i<=n; ++i) for(K=_v; k>=c[i];--k) for(j=1; j<=num[i]&&j*c[i]<=k; ++J)//It's a loop more than a 01 backpack.{TC=j*C[i]; TV=j*V[i]; TEM=dp[k-tc]+TV; if(Tem>dp[k]) dp[k]=tem; }        //printf"%d\n", Dp[_v]); }    return 0;}

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Idea 2: The item spreads out and then handles 01 backpacks (the equivalent of a line filled with items across the line,

2.1 Simple split, the expansion of each of the items, both n pieces a items, the N-piece whether to select the operation

Code 2:JK

#include <iostream>#include<stdio.h>#include<string.h>using namespacestd;intdp[ the];intMain () {intI,j,k,tem; intT,n,_v;//number of test cases, item type, backpack size    intc[ the],v[ the];//Cost, Value    intnum[ the];//number of items per item    intTC,TV;//cost of Split items, valuescanf"%d",&t);  while(t--) {scanf ("%d%d",&_v,&N); Memset (DP,0,sizeof(DP));  for(i=1; i<=n; ++i) scanf ("%d%d%d",&c[i],&v[i],&Num[i]); //////         for(i=1; i<=n; ++i) for(j=1; j<=num[i]&&j*c[i]<=_v; ++J)//It's a loop more than a 01 backpack.                 for(K=_v; k>=j*c[i];--k) {tem=dp[k-c[i]]+V[i]; if(Tem>dp[k]) dp[k]=tem; }        //printf"%d\n", Dp[_v]); }    return 0;}

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hint:2.1 Code and Ideas 1 code is just a different loop order, the actual operation is the same. (and the complexity of the time is the same, slightly troublesome.)
Because
Idea 1:
for (K=_v; k>=c[i];--k)
for (j=1; j<=num[i]&&j*c[i]<=k; ++j)
Idea 2:
for (j=1; j<=num[i]&&j*c[i]<=_v; ++j)
for (K=_v; k>=j*c[i];--k)

Take a look at these two loops,
C[i] K _v
----------------------
1 |   .   .   .    . . |
|   .   .    . . |
J |   .    . . |
|    . . |
m | . |
---------------------()

The cycles of the two loops are the same and are all in the matrix '. ' , that is, the same number of calculations, the same complexity of time


2.2 Binary split, there are n pieces a goods, then split into 1,2,4,8,..., Q So, that is 1+2+4+...+q=n

Code 3:

#include <iostream>#include<stdio.h>#include<string.h>using namespacestd;intdp[ the];intMain () {intI,j,k,tem; intT,n,_v;//number of test cases, item type, backpack size    intc[ the],v[ the];//Cost, Value    intnum[ the];//number of items per item    intTC,TV;//cost of Split items, valuescanf"%d",&t);  while(t--) {scanf ("%d%d",&_v,&N); Memset (DP,0,sizeof(DP));  for(i=1; i<=n; ++i) scanf ("%d%d%d",&c[i],&v[i],&Num[i]); //////         for(i=1; i<=n; ++i) { for(j=1; j<=num[i]; num[i]=num[i]-j,j=j*2)//It's a loop more than a 01 backpack.{TC=j*c[i];//cost of items after splittingTv=j*v[i];//                 for(K=_v; k>=tc;--k) {tem=dp[k-tc]+TV; if(Tem>dp[k]) dp[k]=tem; }            }            if(num[i]>0)//If there is an item, Num[i] is the equivalent of Q in 1+2+4+...+q{TC=num[i]*C[i]; TV=num[i]*V[i];  for(K=_v; k>=tc;--k) {tem=dp[k-tc]+TV; if(Tem>dp[k]) dp[k]=tem; }            }        }    //printf"%d\n", Dp[_v]); }    return 0;}

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HDU 2191 Mourning 512 Wenchuan earthquake--cherish now, Thanksgiving life (multiple backpack)