Diamond Collector （动态规划）-白红宇

Diamond Collector （动态规划）

阅读量：5830 次

发布时间：2019-06-18

本文共 1884 字，大约阅读时间需要 6 分钟。

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 31051129514

样例输出

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                   #define inf 0x3f3f3f3f#define mod 1000000007typedef 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
                   
                    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
                    
                     =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<
                     
                      <

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转载于:https://www.cnblogs.com/jianrenfang/p/5719856.html

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