CodeForces 444E

本文深入探讨了并查集算法的实现细节,包括关键的数据结构定义、查找与合并操作的具体实现方式,以及如何通过示例代码展示算法的实际应用。特别地,文章中详细解释了如何利用并查集解决特定问题,例如在图论问题中如何有效地进行连通性判断。

并查集

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=3300;
struct Node {
    int u,v,w;
    Node(){}
    Node(int u,int v,int w):u(u),v(v),w(w){}
    bool operator<(const Node & th)const {
        return w<th.w;
    }
}node[N];
int cnt[N];
int fa[N],num[N];
int Find(int u){
    if(u==fa[u])return u;
    return fa[u]=Find(fa[u]);
}

int main(){
    #ifdef DouBi
    freopen("in.cpp","r",stdin);
    #endif // DouBi
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i=0;i<n-1;i++){
            int a,b,c;scanf("%d%d%d",&a,&b,&c);
            node[i]=Node(a,b,c);
        }
        sort(node,node+n-1);
        int sum=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&cnt[i]);
            sum+=cnt[i];
        }
        for(int i=1;i<=n;i++){
            fa[i]=i,num[i]=1;
        }
        int ans=0;
        for(int i=0;i<n-1;i++){
            int u=node[i].u,v=node[i].v,w=node[i].w;
            //printf("%d %d %d\n",u,v,w);
            int x=Find(u),y=Find(v);
            fa[x]=y;
            cnt[y]+=cnt[x];
            num[y]+=num[x];
            ans=w;
            if(sum-cnt[y]<num[y])break;
        }
        printf("%d\n",ans);
    }
    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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