hdu 2660 Accepted Necklace
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=2660
Accepted Necklace
Description
I have N precious stones, and plan to use K of them to make a necklace for my mother, but she won't accept a necklace which is too heavy. Given the value and the weight of each precious stone, please help me find out the most valuable necklace my mother will accept.
Input
The first line of input is the number of cases.
For each case, the first line contains two integers N (N <= 20), the total number of stones, and K (K <= N), the exact number of stones to make a necklace.
Then N lines follow, each containing two integers: a (a<=1000), representing the value of each precious stone, and b (b<=1000), its weight.
The last line of each case contains an integer W, the maximum weight my mother will accept, W <= 1000.
Output
For each case, output the highest possible value of the necklace.
Sample Input
1
2 1
1 1
1 1
3
Sample Output
1
dfs。。
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<set>
using std::max;
using std::sort;
using std::pair;
using std::swap;
using std::queue;
using std::multiset;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 30;
const int INF = 0x3f3f3f3f;
typedef unsigned long long ull;
struct Node {
int v, w;
}A[N];
bool vis[N];
int W, K, n, ans;
void dfs(int cur, int w, int v, int tot) {
if (tot == K) {
ans = max(ans, v);
return;
}
for (int i = cur; i < n; i++) {
if (!vis[i] && tot + 1 <= K && w + A[i].w <= W) {
vis[i] = true;
dfs(i + 1, w + A[i].w, v + A[i].v, tot + 1);
vis[i] = false;
}
}
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
int t;
scanf("%d", &t);
while (t--) {
ans = -INF;
scanf("%d %d", &n, &K);
rep(i, n) {
vis[i] = false;
scanf("%d %d", &A[i].v, &A[i].w);
}
scanf("%d", &W);
dfs(0, 0, 0, 0);
printf("%d\n", ans);
}
return 0;
}
hdu 2660 Accepted Necklace的更多相关文章
- HDOJ(HDU).2660 Accepted Necklace (DFS)
HDOJ(HDU).2660 Accepted Necklace (DFS) 点我挑战题目 题意分析 给出一些石头,这些石头都有自身的价值和重量.现在要求从这些石头中选K个石头,求出重量不超过W的这些 ...
- HDU 2660 Accepted Necklace【数值型DFS】
Accepted Necklace Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others ...
- hdu 2660 Accepted Necklace(dfs)
Problem Description I have N precious stones, and plan to use K of them to make a necklace for my mo ...
- hdu - 2660 Accepted Necklace (二维费用的背包问题)
http://acm.hdu.edu.cn/showproblem.php?pid=2660 f[v][u]=max(f[v][u],f[v-1][u-w[i]]+v[i]; 注意中间一层必须逆序循环 ...
- Accepted Necklace
Accepted Necklace Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) T ...
- hdu 5730 Shell Necklace [分治fft | 多项式求逆]
hdu 5730 Shell Necklace 题意:求递推式\(f_n = \sum_{i=1}^n a_i f_{n-i}\),模313 多么优秀的模板题 可以用分治fft,也可以多项式求逆 分治 ...
- hdu2660 Accepted Necklace (DFS)
Problem Description I have N precious stones, and plan to use K of them to make a necklace for my mo ...
- HDU 5730 Shell Necklace(CDQ分治+FFT)
[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5730 [题目大意] 给出一个数组w,表示不同长度的字段的权值,比如w[3]=5表示如果字段长度为3 ...
- hdu 5730 Shell Necklace——多项式求逆+拆系数FFT
题目:http://acm.hdu.edu.cn/showproblem.php?pid=5730 可以用分治FFT.但自己只写了多项式求逆. 和COGS2259几乎很像.设A(x),指数是长度,系数 ...
随机推荐
- 转载 《AngularJS》5个实例详解Directive(指令)机制
<AngularJS>5个实例详解Directive(指令)机制 大漠穷秋 本文整理并扩展了<AngularJS>这本书第六章里面的内容,此书近期即将由电子工业出版社出版,敬请 ...
- 深入了解Qt(三)之元signal和slot
深入了解Qt主要内容来源于Inside Qt系列,本文做了部分删改,以便于理解.在此向原作者表示感谢! 在Qt 信号和槽函数这篇文章中已经详细地介绍了信号和槽的使用及注意事项.在这里对其使用方面的知识 ...
- poj1005 I Think I Need a Houseboat
这题目只要读懂了意思就好做了,先求出来(0.0)到(x.y)的距离为r,然后求出来以r为半径的半圆的面积,然后再用这个面积除以50,再向上取整就可以啦. #include <stdio.h> ...
- Android IOS WebRTC 音视频开发总结(三九)-- win10升级为何要p2p
本文主要介绍webrtc p2p的应用场景,文章来自博客园RTC.Blacker,支持原创,转载请说明出处. P2P最简单的解释就是两个客户端之间直接进行数据交互,不经过服务端转发. 最早接触P2P是 ...
- python简单的发送邮件
python 利用smtplib来发送邮件,具体的代码如下 一. 编辑smtp_v2.py vim /home/python/smtp_v2.py #!/usr/bin/env python # -* ...
- 2_JavaScript日期格式化
第二章 JavaScript 时间格式化 2.1 Ticks 转换为常规日期 2.2 常规日期格式化 <input type="button" value=&qu ...
- luigi学习2-在hadoop上运行Top Artists
一.AggregateArtistsHadoop class AggregateArtistsHadoop(luigi.contrib.hadoop.JobTask): date_interval = ...
- 在Activity中设置new出来的TextView属性
//创建一个TextView---->textView TextView textView = new TextView(this); // 第一个参数为宽的设置,第二个参数为高的设置 te ...
- jquery实现radio按纽全不选和checkbox全选的实例
用jquery实现以下两个这个功能: 1.对所有单选按纽中radio全不选 单选按纽:<input type="radio" name="f1">A ...
- DevExpress GridControl 部分用法
1.GridControl赋值:this.GridControl1.DataSouce=dt; 2.GridContro总合计及分组合计: 常规总合计直接RunDesigner-Group Summa ...