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  • poj 2754 Similarity of necklaces 2 夜

    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;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/liulangye/p/2917216.html
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