Codeforces 1004E

本文解析了CodeForces竞赛中一道关于树的直径及滑动窗口算法的问题。通过两次深度优先搜索找到树的直径,并利用滑动窗口优化路径长度,最终求得树上任一点到其它所有点的最大距离。

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题目链接:http://codeforces.com/contest/1004/problem/E

题目描述:

 

 

题意:
 简单来说就是给你一颗n个节点的树,然后你自己选择一条长度不超过k个节点的路径(必须是一条树上的路径)。将这条路径看成一个点,问这个点到树上其他点的最远距离是多少? 

题解:

首先找到树的一条直径(两次DFS)。并且处理出直径起点到其他点的距离。 

 然后再这个直径上做滑动窗口。滑动窗口的长度不超过k。取窗口到直径两端距离较大的那个。最后的ans和它取min。 
 然后在O(n)遍历的时候,dfs找出这个点到其他叶子的最远距离,ans和它取max。注意,这个dfs的过程是到其他叶子的最远距离,若碰到直径上的点就return。

代码:

#include <bits/stdc++.h>

using namespace std;
#define fi first
#define se second
typedef long long ll;
const int maxn = 100007;
vector< pair<int, int> > g[maxn];
vector<int> diameter;
ll dep[maxn];
ll dis[maxn];
int par[maxn];
int n, k;
int st, ed;
bool vis[maxn];

void dfs1(int u, int f)
{
    par[u] = f;
    for(auto it : g[u]) {
        int v = it.fi;
        int d = it.se;
        if(v != f) {
            dis[v] = dis[u] + d;
            dfs1(v, u);
        }
    }
}

void dfs2(int u, int f) //寻找直径上的点到其他非直径上的点的最长距离
{
    for(auto it : g[u]) {
        if(!vis[it.fi] && it.fi != f) {
            dfs2(it.fi, u);
            dep[u] = max(dep[u], dep[it.fi] + it.se); //注意是回溯的过程
        }
    }
}

void findDiameter() //两次DFS寻找直径
{
    st = 1, ed = 1;
    dis[st] = 0;
    dfs1(st, -1);
    int maxd = 0;
    for(int i = 1; i <= n; i ++) {
        if(dis[i] > maxd) {
            st = i;
            maxd = dis[i];
        }
    }
    memset(dis, 0, sizeof(dis));
    dfs1(st, -1);
    maxd = dis[st];
    for(int i = 1; i <= n; i ++) {
        if(dis[i] > maxd) {
            ed = i;
            maxd = dis[i];
        }
    }
    for(int i = ed; i != -1; i = par[i]) {
        diameter.push_back(i); // 将直径上的点存储下来
        vis[i] = 1; //进行标记
    }
//    for(auto it : diameter) cout << it << ' ';
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0);

    cin >> n >> k;
    for(int i = 0; i < n - 1; i ++) {
        int u, v, w;
        cin >> u >> v >> w;
        g[u].push_back(make_pair(v, w));
        g[v].push_back(make_pair(u, w));
    }
    findDiameter();
    ll res = 1LL << 59;
    ll hmax = 0;
    for(int i = 0; i < diameter.size(); i ++) {
        ll x = 0;
        if(i >= k) {
            x = dis[ed] - dis[diameter[i - k + 1]]; // 当前diameter[i]点到直径末尾的距离
        }
        x = max(x, dis[diameter[i]]);  // 当前diameter[i]点到直径开始的距离
        dfs2(diameter[i], -1); //寻找当前点到非直径点的最长距离
        hmax = max(hmax, dep[diameter[i]]); //hmax需要一直保留,当窗口进行滑动使,可能有影响.
        res = min(res, max(x, hmax));
    }
    cout << res << '\n';
    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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