[Codeforces 125E] MST Company (单度限制最小生成树)

该博客探讨了Codeforces 125E问题,涉及构建一棵边权最小的树,同时确保起点1的度数为k。区别于传统的度限制最小生成树问题,此题要求根节点度数严格等于k。作者提出使用二分法结合Kruskal算法来解决,通过对与根节点连接的边进行预处理,调整边的权重以满足条件,并在二分过程中计算与节点1相连的边的数量。

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题意:

有N个点,M条边,要生成一颗树。使得边权最小,而且对于起点 1 有度限制(k)。如果可以,输出所选边,否则输出 -1.


这道题可以用度限制最小生成树来做,但是这道题和度限制最小生成树又有区别,这里要求根节点的度必须为K。而后者要求小于等于K的最小值。

所以这道题目可以采用二分来实现。

采用krusical来做最小生成树的时候,每次挑选最小可行边加入集合,所以可以预处理一下与根节点相连的边,变为 w[i] ' = mid + w[i]. 

通过二分mid,每次求最小生成树的时候记录与1结点相连的边数。


/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define maxn 5500
#define maxm 100100
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define For(i, n) for (int i = 0; i < n; i++)
typedef long long ll;
using namespace std;
int n, m, k, x[maxm], y[maxm], w[maxm], p[m
### 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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