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

  1. 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 ...

  2. HDU 2660 Accepted Necklace【数值型DFS】

    Accepted Necklace Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others ...

  3. 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 ...

  4. 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 ...

  5. Accepted Necklace

    Accepted Necklace Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) T ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. oracle 索引失效原因

    转自  http://www.cnblogs.com/orientsun/archive/2012/07/05/2577351.html Oracle 索引的目标是避免全表扫描,提高查询效率,但有些时 ...

  2. 华为OJ平台——计算字符串的相似度

    题目描述: 对于不同的字符串,我们希望能有办法判断相似程度,我们定义了一套操作方法来把两个不相同的字符串变得相同,具体的操作方法如下: 1 修改一个字符,如把“a”替换为“b”. 2 增加一个字符,如 ...

  3. java基础回顾(一)—— sleep和wait的区别

    sleep是Thread类的一个方法,wait是Object类的一个方法 sleep是线程被调用时,占着cpu去睡觉,其他线程不能占用cpu,os认为该线程正在工作,不会让出系统资源 wait是进入等 ...

  4. 【MySQL】SQL语句嵌套1

    mysql中You can't specify target table <tbl> for update in FROM clause错误的意思是说,不能先select出同一表中的某些值 ...

  5. C# 共用的返回数据类

    using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Dscf ...

  6. Windows phone 8 学习笔记(7) 设备(转)

    本节主要涉及到 Windows phone 8 手机支持的各类设备,包括相机.设备状态,振动装置等.还有各类感应器,包括磁力计.加速度器和陀螺仪.通过设备状态可以获取内存.硬件.电源.键盘等状态:通过 ...

  7. WeChat 6.3 wipe deleted chat messages as well as LINE 5.3 and above

    Let me show you the WeChat version first. It is 6.3. What will happen to WeChat deleted chat message ...

  8. IIS报错 未将对象引用设置到对象的实例。

    在vs中运行正常的项目 ,发布到IIS总是提示 未将对象引用设置到对象的实例. 运行静态页面 html正常,只是打开.aspx页面的时候报错,在确保了数据库,配置,权限均正常的情况下. 错误原因:先安 ...

  9. C++ inline(内联什么时候使用)

    (1)什么是内联函数?内联函数是指那些定义在类体内的成员函数,即该函数的函数体放在类体内. (2)为什么要引入内联函数?当然,引入内联函数的主要目的是:解决程序中函数调用的效率问题.另外,前面我们讲到 ...

  10. ADO.NET中的Connection详解

    连接字符串 1.写法一 "Data Source=服务器名; Initial Catalog=数据库; User ID =用户名; Password=密码; Charset=UTF8; &q ...