Diamond Collector （动态规划）

问题 I: Diamond Collector

时间限制: 1 Sec  内存限制: 64 MB
提交: 22  解决: 7
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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

7 3
10
5
1
12
9
5
14

样例输出

5

#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=;
int n,dp[N],len,g[N];
ll k,a[N];
int w[][];
int flag=;
int vis[]={};
string str[],ch;
int maxn=;
map<string,int>p,pp;

int main() {
    memset(dp,,sizeof(dp));
    cin>>n>>k;
    for(int i=;i<=n;i++)cin>>a[i];
    sort(a+,a+n+);
    int m=,r=;
    for(int i=;i<=n;i++)
    {
       while(a[r+]-a[i]<=k&&r<n) r++;
       dp[i]=r-i+;
    }
    for(int i=n;i>=;i--)
        g[i]=max(g[i+],dp[i]);
    int ans=;
    for(int i=;i<=n;i++)
        ans=max(ans,dp[i]+g[i+dp[i]]);
    cout<<ans<<endl;
    return ;
}