AC日记——[USACO07DEC]手链Charm Bracelet 洛谷 P2871

题目描述

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.

有N件物品和一个容量为V的背包。第i件物品的重量是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重量总和不超过背包容量,且价值总和最大。

输入输出格式

输入格式:

  • 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

输出格式:

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

输入输出样例

输入样例#1:
4 6
1 4
2 6
3 12
2 7
输出样例#1:
23


思路:
  裸01背包;


来,上代码:
#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

int if_z,dp[15005],n,m,vi,ci;

char Cget;

inline void in(int &now)
{
    now=0,if_z=1,Cget=getchar();
    while(Cget>'9'||Cget<'0')
    {
        if(Cget=='-') if_z=-1;
        Cget=getchar();
    }
    while(Cget>='0'&&Cget<='9')
    {
        now=now*10+Cget-'0';
        Cget=getchar();
    }
    now*=if_z;
}

int main()
{
    in(n),in(m);
    while(n--)
    {
        in(vi),in(ci);
        for(int i=m;i>=vi;i--) dp[i]=max(dp[i],dp[i-vi]+ci);
    }
    cout<<dp[m];
    return 0;
}
原文地址:https://www.cnblogs.com/IUUUUUUUskyyy/p/6485815.html