Diamond Collector （动态规划）

最新推荐文章于 2025-05-16 07:15:40 发布

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原文链接：http://www.cnblogs.com/jianrenfang/p/5719856.html

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#数据结构与算法

本文介绍了一道关于钻石收集的算法题，旨在寻找最优方案来展示不同大小的钻石。问题要求将钻石分配到两个展柜中，使得每个展柜内的钻石大小相近，且总体展示的钻石数量最多。

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问题 I: Diamond Collector

时间限制: 1 Sec 内存限制: 64 MB
提交: 22 解决: 7
[提交][状态][讨论版]

题目描述

Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare time! She has collected N diamonds (N≤50,000) of varying sizes, and she wants to arrange some of them in a pair of display cases in the barn.

Since Bessie wants the diamonds in each of the two cases to be relatively similar in size, she decides that she will not include two diamonds in the same case if their sizes differ by more than K
(two diamonds can be displayed together in the same case if their sizes differ by exactly K). Given K, please help Bessie determine the maximum number of diamonds she can display in both cases together.

输入

The first line of the input file contains N and K (0≤K≤1,000,000,000). The next N lines each contain an integer giving the size of one of the diamonds. All sizes will be positive and will not exceed 1,000,000,000.

输出

Output a single positive integer, telling the maximum number of diamonds that Bessie can showcase in total in both the cases.

样例输入

样例输出

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 1000000007
typedef long long ll;
using namespace std;
const int N=50010;
int n,dp[N],len,g[N];
ll k,a[N];
int w[21][21];
int flag=0;
int vis[10000]={0};
string str[21],ch;
int maxn=1;
map<string,int>p,pp;
 
 
int main() {
    memset(dp,0,sizeof(dp));
    cin>>n>>k;
    for(int i=1;i<=n;i++)cin>>a[i];
    sort(a+1,a+n+1);
    int m=1,r=1;
    for(int i=1;i<=n;i++)
    {
       while(a[r+1]-a[i]<=k&&r<n) r++;
       dp[i]=r-i+1;
    }
    for(int i=n;i>=1;i--)
        g[i]=max(g[i+1],dp[i]);
    int ans=0;
    for(int i=1;i<=n;i++)
        ans=max(ans,dp[i]+g[i+dp[i]]);
    cout<<ans<<endl;
    return 0;
}