题目传送门:P2871 [USACO07DEC]手链Charm Bracelet

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

第一行:物品个数N和背包大小M

第二行至第N+1行:第i个物品的重量C[i]和价值W[i]

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

输出一行最大价值。

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

题解:

  01背包模板题。

#include<bits/stdc++.h>
using namespace std;
const int N = ;
const int M = ;
int c[N],w[N],dp[M]={};
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for (int i=;i<=n;i++)
scanf("%d%d",&c[i],&w[i]);
for (int i=;i<=n;i++)
for (int j=m;j>=c[i];j--)
dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
printf("%d\n",dp[m]);
return ;
} 01背包

P2871 [USACO07DEC]手链Charm Bracelet(01背包模板)的更多相关文章

  1. 洛谷 P2871 [USACO07DEC]手链Charm Bracelet && 01背包模板

    题目传送门 解题思路: 一维解01背包,突然发现博客里没有01背包的板子,补上 AC代码: #include<cstdio> #include<iostream> using ...

  2. 洛谷——2871[USACO07DEC]手链Charm Bracelet——01背包

    题目描述 Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like t ...

  3. bzoj1625 / P2871 [USACO07DEC]手链Charm Bracelet

    P2871 [USACO07DEC]手链Charm Bracelet 裸01背包. 看到自己1年半前写的30分code.......菜的真实(捂脸) #include<iostream> ...

  4. POJ 3624 Charm Bracelet(01背包模板题)

    题目链接 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 52318   Accepted: 21912 Descriptio ...

  5. 洛谷 P2871 [USACO07DEC]手链Charm Bracelet 题解

    题目传送门 这道题明显就是个01背包.所以直接套模板就好啦. #include<bits/stdc++.h> #define MAXN 30000 using namespace std; ...

  6. P2871 [USACO07DEC]手链Charm Bracelet

    题目描述 Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like t ...

  7. 洛谷——P2871 [USACO07DEC]手链Charm Bracelet

    https://www.luogu.org/problem/show?pid=2871 题目描述 Bessie has gone to the mall's jewelry store and spi ...

  8. 【做题笔记】P2871 [USACO07DEC]手链Charm Bracelet

    就是 01 背包.大意:给您 \(T\) 个空间大小的限制,有 \(M\) 个物品,第 \(i\) 件物品的重量为 \(c_i\) ,价值为 \(w_i\) .要求挑选一些物品,使得总空间不超过 \( ...

  9. 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 t ...

随机推荐

  1. div盒子或者图片并排居中

    要使div总是找不到原因居中很简单,float和display都可以实现,float就不说了,这里说一下display:line-block,比如四个或者多个div盒子,明明设置好了宽度后,总有一个上 ...

  2. poj 3335 Rotating Scoreboard (Half Plane Intersection)

    3335 -- Rotating Scoreboard 给出一个多边形,要求判断它的内核是否存在. 还是半平面交的题,在这道题中,公告板允许其所在位置与直线共线也算是可见,于是我们就可以将每一条直线微 ...

  3. 【原生JS】简单取随机数

    因为js本身取随机数仅能取 0 到 1之间的数,所以..... 取 1 - 10之间的随机数. function getSJS(x,y,z){ var int = null; while(int &g ...

  4. npm镜像及配置方法

    npm全称Node Package Manager,是node.js的模块依赖管理工具.由于npm的源在国外,所以国内用户使用起来各种不方便.下面整理出了一部分国内优秀的npm镜像资源,国内用户可以选 ...

  5. canvas实现碰壁反弹(单个小方块)

    <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  6. Adam那么棒,为什么还对SGD念念不忘 (1) —— 一个框架看懂优化算法

    机器学习界有一群炼丹师,他们每天的日常是: 拿来药材(数据),架起八卦炉(模型),点着六味真火(优化算法),就摇着蒲扇等着丹药出炉了. 不过,当过厨子的都知道,同样的食材,同样的菜谱,但火候不一样了, ...

  7. css的一些小问题

    这是今天整理的笔记一.属性书写顺序: Formatting Model(布局方式.位置) > Box Model(尺寸) > Typographic(文本相关) > Visual(视 ...

  8. H3C RIP基本配置

  9. H3C RIP路由表的更新

  10. H3C RIP协议概述