codeforces 1154E Two Teams

本文解析了TwoTeams问题的算法解决方案,通过使用C++ STL中的list进行高效操作,实现了两个教练挑选队员进入各自集训队的过程。算法确保每次选择分数最高者及其周边最多k人,并标记已选人员,最终确定每位选手所属队伍。

题目链接:Two Teams

题意:有两个教练,team=1和team=2,两个教练分别要挑人进入自己的集训队。每次挑分数最高的以及他两边最多k个人进队伍。问挑完以后哪些人进了1队哪些人进了2队

思路:在网上看到了大佬们用c++的STL里的list进行操作,即可

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<stdlib.h>
#include<map>
#include<set>
#include<vector>
#include<memory.h>
#include<list>
using namespace std;
const int maxn = 2 * 100000 + 10;
typedef long long ll;
list<pair<int, int> > L;
list<pair<int, int> >::iterator ite[maxn];
pair<int, int> a[maxn];
bool vis[maxn];
int ans[maxn];
int n, k;
int main()
{
	ios::sync_with_stdio(false);
	cin >> n >> k;
	int team = 1;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i].first;
		a[i].second = i;
		L.push_back(a[i]);
		ite[i] = --L.end();
	}
	sort(a + 1, a + 1 + n);
	for (int i = n; i >= 1; i--)
	{
		int id = a[i].second;
		if (vis[id])
			continue;
		list<pair<int, int> >::iterator l = ite[id], r = ite[id];
		for (int i = 1; i <= k; i++)
		{
			if (l != L.begin())
				l--;
			if (r != --L.end())
				r++;
		}
		r++;
		while (l != r)
		{
			ans[l->second] = team;
			vis[l->second] = true;
			l = L.erase(l);
		}
		team = 3 - team;
	}
	for (int i = 1; i <= n; i++)
		cout << ans[i];
	return 0;
}

 

 

### 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?
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