Diamond Collector

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

5分析：取两个盒子，每个盒子里的数的极差不超过k，问两个盒子里的数最大的和是多少；　　　排序，预处理每个数能达到的最大长度；　　　从后向前更新最大值，注意每次更新完之后，可以把当前数达到的最大长度更新；代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=1e5+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
using namespace __gnu_cxx;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,a[maxn],p[maxn],ma;
int main()
{
    int i,j,k,t;
    scanf("%d%d",&n,&k);
    rep(i,0,n-1)scanf("%d",&a[i]);
    sort(a,a+n);
    rep(i,0,n-1)
    {
        int l=i,r=n-1,ans=i;
        while(l<=r)
        {
            int mid=l+r>>1;
            if(a[mid]<=a[i]+k)ans=mid,l=mid+1;
            else r=mid-1;
        }
        p[i]=ans-i+1;
    }
    for(i=n-1;i>=0;i--)
    {
        ma=max(ma,p[i]+p[i+p[i]]);
        p[i]=max(p[i],p[i+1]);
    }
    printf("%d\n",ma);
    //system ("pause");
    return 0;
}

时间： 2024-10-02 09:15:43

Diamond Collector