Codeforces 985E

本文介绍了一种利用动态规划和单调队列优化的方法来解决排序后的数组能否成功按要求打包的问题,并提供了完整的代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题意略。

思路:

这个题目开始想的有点暴力,后来发现有搜索的性质,因此转而用动态规划。首先,我们要把这些数排个序。

定义状态:dp[i]为排序后i~n能否成功打包,1表示可以,0表示不能打包。

状态转移方程:dp[i] = max{dp[j]} (i + k <= j <= upper)upper为在有序数组中,第一个比当前值store[i] + d大的数的下标。

为了防止它卡数据,这里可以用单调队列优化一下,由于确定upper要使用二分搜索,所以总的复杂度是O(nlogn)。

详见代码:

#include<bits/stdc++.h>
#define maxn 500005
using namespace std;

int dp[maxn],que[maxn],head,tail,store[maxn];
int n,k,d;

int main(){
    scanf("%d%d%d",&n,&k,&d);
    for(int i = 1;i <= n;++i) scanf("%d",&store[i]);
    sort(store + 1,store + 1 + n);
    dp[n + 1] = 1;
    for(int i = n;i >= 1;--i){
        int temp = store[i] + d;
        int r = upper_bound(store + 1,store + 1 + n,temp) - store;
        int l = i + k;
        if(l > r){
            dp[i] = 0;
        } 
        else{
            while(que[head] > r && head < tail) ++head;
            while(head < tail && dp[que[tail - 1]] < dp[l]) --tail;
            que[tail++] = l;
            dp[i] = dp[que[head]];
        }
    }
    printf("%s\n",dp[1] ? "YES" : "NO");
    return 0;
}

/*
18 3 1
1 1 1 2 2 3 5 5 5 6 6 7 9 9 9 10 10 11

YES
*/

 

转载于:https://www.cnblogs.com/tiberius/p/9162380.html

### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值