多重背包

题目大意：有N种物品和一个容量为V的背包。第i种物品最多有num[i]件可用，每件费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量，且价值总和最大。

状态转移方程为：f[i][v]=max{f[i-1][v-k*c[i]]+k*w[i]|0<=k<=n[i]};

#include <iostream>

#include <string.h>

#include <stdio.h>

using namespace std;

const int N = 1005;

int dp
;

int c
,w
,num
;

int n,m;

void ZeroOne_Pack(int cost,int weight,int n)

{

for(int i=n; i>=cost; i--)

dp[i] = max(dp[i],dp[i-cost] + weight);

}

void Complete_Pack(int cost,int weight,int n)

{

for(int i=cost; i<=n; i++)

dp[i] = max(dp[i],dp[i-cost] + weight);

}

int Multi_Pack(int c[],int w[],int num[],int n,int m)

{

memset(dp,0,sizeof(dp));

for(int i=1; i<=n; i++)

{

if(num[i]*c[i] > m)

Complete_Pack(c[i],w[i],m);

else

{

int k = 1;

while(k < num[i])

{

ZeroOne_Pack(k*c[i],k*w[i],m);

num[i] -= k;

k <<= 1;

}

ZeroOne_Pack(num[i]*c[i],num[i]*w[i],m);

}

}

return dp[m];

}

int main()

{

int t;

cin>>t;

while(t--)

{

cin>>m>>n;

for(int i=1; i<=n; i++)

cin>>c[i]>>w[i]>>num[i];

cout<<Multi_Pack(c,w,num,n,m)<<endl;

}

return 0;

}