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  • poj 3624 Charm Bracelet 背包DP

    Charm Bracelet

    Time Limit: 1 Sec  Memory Limit: 256 MB

    题目连接

    http://poj.org/problem?id=3624

    Description

    Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

    Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

    Input

    * Line 1: Two space-separated integers: N and M
    * Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di

    Output

    * Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

    Sample Input

    4 6
    1 4
    2 6
    3 12
    2 7

    Sample Output

    23

    HINT

    题意

    01背包裸题

    题解:

    01背包,滚动数组优化一下

    代码:

    //qscqesze
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    #include <map>
    #include <stack>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 200001
    #define mod 10007
    #define eps 1e-9
    int Num;
    char CH[20];
    //const int inf=0x7fffffff;   //нчоч╢С
    const int inf=0x3f3f3f3f;
    /*
    
    inline void P(int x)
    {
        Num=0;if(!x){putchar('0');puts("");return;}
        while(x>0)CH[++Num]=x%10,x/=10;
        while(Num)putchar(CH[Num--]+48);
        puts("");
    }
    */
    //**************************************************************************************
    inline ll read()
    {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    inline void P(int x)
    {
        Num=0;if(!x){putchar('0');puts("");return;}
        while(x>0)CH[++Num]=x%10,x/=10;
        while(Num)putchar(CH[Num--]+48);
        puts("");
    }
    
    struct node
    {
        int v,vl;
    };
    node a[maxn];
    int dp[maxn];
    int main()
    {
        int n,v;
        while(scanf("%d%d",&n,&v)!=EOF)
        {
            memset(a,0,sizeof(a));
            memset(dp,0,sizeof(dp));
            //n=read(),v=re9ad();
            for(int i=1;i<=n;i++)
                a[i].vl=read(),a[i].v=read();
            for(int i=1;i<=n;i++)
            {
                for(int j=v;j>=a[i].vl;j--)
                {
                    dp[j]=max(dp[j],dp[j-a[i].vl]+a[i].v);
                }
            }
            cout<<dp[v]<<endl;
        }
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4455678.html
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