poj 2754 Similarity of necklaces 2

http://poj.org/problem?id=2754

先把low--up 转化为0--(up-low)

然后变成背包  背包的关键在于多重背包用二进制优化

代码：

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<map>

#include<vector>

#include<stack>

#include<set>

#include<map>

#include<queue>

#include<algorithm>

#include<cmath>

#define LL long long

//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const int INF=0x3f3f3f3f;

const int N=205;

const int M=100005;

int low[N],up[N],p[N],m[N];

int dp[M];

void pack01(int cost,int weight,int V)

{

    for(int v=V;v>=cost;--v)

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

}

void packComplete(int cost,int weight,int V)

{

    for(int v=cost;v<=V;++v)

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

}

void packMultiple(int cost,int weight,int num,int V)

{

    if((num+1)*cost>V)

    packComplete(cost,weight,V);

    else

    {

        int k=1;

        while(k<=num)

        {

            pack01(cost*k,weight*k,V);

            num-=k;

            k=k<<1;

        }

        pack01(cost*num,weight*num,V);

    }

}

int main()

{

    //freopen("data.in","r",stdin);

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        int k=0,res=0;

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

        {

            scanf("%d %d %d %d",&p[i],&m[i],&low[i],&up[i]);

            k+=(0-low[i])*m[i];

            res+=(low[i]-0)*p[i];

        }

        memset(dp,0x80,sizeof(dp));

        dp[0]=0;

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

        packMultiple(m[i],p[i],up[i]-low[i],k);

        printf("%d\n",dp[k]+res);

    }

    return 0;

}