[转]POJ3624 Charm Bracelet(典型01背包问题)
来源:https://www.cnblogs.com/jinglecjy/p/5674796.html
题目链接:http://bailian.openjudge.cn/practice/4131/
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 32897 Accepted: 14587
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
一、题目大意
有N个物品,分别有不同的重量Wi和价值Di,Bessie只能带走重量不超过M的物品,要是总价值最大,并输出总价值。
二、解题思路
典型的动态规划题目,用一个数组记录背包各个重量的最优解,不断地更新直到穷尽所有可能性。
状态更新公式:state[weight] = max{state[weight-W[i]]+D[i], state[weight]}
// 背包问题(动态规划)
#include <iostream>
#include <cstdio>
#include <cstring>
#define MAXN 3402
#define MAXM 12880
using namespace std; int main(){
int N, M, W[MAXN+], D[MAXN+], dp[MAXM+];
while(scanf("%d%d", &N, &M) != EOF){
for(int i=; i<N; i++){
scanf("%d%d", &W[i], &D[i]);
}
memset(dp, , sizeof(dp));
for(int i=; i<N; i++){
for(int left_w=M; left_w>=W[i]; left_w--){
dp[left_w] = max(dp[left_w-W[i]]+D[i], dp[left_w]);
}
}
printf("%d\n", dp[M]);
} return ;
}
[转]POJ3624 Charm Bracelet(典型01背包问题)的更多相关文章
- Charm Bracelet(01背包问题)
题目链接: https://vjudge.net/problem/POJ-3624 题目描述: Bessie has gone to the mall's jewelry store and spie ...
- Poj3624 Charm Bracelet (01背包)
题目链接:http://poj.org/problem?id=3624 Description Bessie has gone to the mall's jewelry store and spie ...
- POJ3624 Charm Bracelet 【01背包】
Charm Bracelet Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22621 Accepted: 10157 ...
- POJ.3624 Charm Bracelet(DP 01背包)
POJ.3624 Charm Bracelet(DP 01背包) 题意分析 裸01背包 代码总览 #include <iostream> #include <cstdio> # ...
- poj3624 Charm Bracelet(DP,01背包)
题目链接 http://poj.org/problem?id=3624 题意 有n个手镯,每个手镯有两个属性:重量W,需求因子D.还有一个背包,它能装下总重量不超过M的手镯.现在将一些镯子装入背包,求 ...
- POJ 3624 Charm Bracelet(01背包模板)
Charm Bracelet Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 45191 Accepted: 19318 ...
- poj 3524 Charm Bracelet(01背包)
Description Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd ...
- POJ 3624 Charm Bracelet (01背包)
题目链接:http://poj.org/problem?id=3624 Bessie has gone to the mall's jewelry store and spies a charm br ...
- Charm Bracelet(01背包)
Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fil ...
随机推荐
- 《剑指offer》-统计整数二进制表示中1的个数
题目描述 输入一个整数,输出该数二进制表示中1的个数.其中负数用补码表示. 直观思路就是把二进制表示从右往左统计1的个数.直接想到移位操作来迭代处理.坑点在于负数的移位操作会填充1.有人贴出了逻辑移位 ...
- [转] React 是什么
用脚本进行DOM操作的代价很昂贵.有个贴切的比喻,把DOM和JavaScript各自想象为一个岛屿,它们之间用收费桥梁连接,js每次访问DOM,都要途径这座桥,并交纳“过桥费”,访问DOM的次数越多, ...
- [转] Web MVC简介
http://blog.csdn.net/zk_software/article/details/8141843
- vtiger7新模块的创建和配置
vtiger出7.0了,以前的那些配置方法已经不管用了 下面是新的 模块创建及一些页面及功能配置的方法 下面介绍三个点 1.新建一个模块 2.实现单图片上传的功能 3.实现页面summary显示的功能 ...
- 【开源小软件 】Bing每日壁纸 V1.2.1
Bing每日壁纸发布V1.2版本,下载地址Release V1.2.1 该小软件可以自动获取Bing的精美图片设置为壁纸,并且支持随机切换历史壁纸,查看壁纸故事. 本次新增国际化支持,以及桌面widg ...
- 配置多个数据源,spring profile 多环境配置管理
针对生产环境,测试环境,以及本地调试开发有时会配置多套数据库,在一个数据配置文件进行修改,往往有时发布到生成环境会忘记修改,或者本地调试时还是生产环境的库,会导致生产环境数据被污染. ps--刚开始配 ...
- input禁止输入空格
<input name="" onkeyup="this.value=this.value.replace(/^\s+|\s+$/g,'')" value ...
- centos7 安装步骤
这里选择64位 32位没有找到网卡... 注:这里是网络类型分配,网络类型分配分为三种,Bridge,NAT和Host-Only,大概区别是 1 BRIDGE 桥接:相当于主机和虚拟机连接到同一个h ...
- windows下端口映射(端口转发)
windows下端口映射(端口转发) 转载: https://blog.csdn.net/i1j2k3/article/details/70228043 本文是对网文的归纳整理,算不上原创,摸索过程亲 ...
- 【Java并发核心四】Executor 与 ThreadPoolExecutor
Executor 和 ThreadPoolExecutor 实现的是线程池,主要作用是支持高并发的访问处理. Executor 是一个接口,与线程池有关的大部分类都实现了此接口. ExecutorSe ...