Submission #875414

#	Time	Username	Problem	Language	Result	Execution time	Memory
875414		AverageAmogusEnjoyer	Dreaming (IOI13_dreaming)	C++17	18 / 100	45 ms	19420 KiB

dreaming

#include "bits/stdc++.h"
#include "dreaming.h"
using namespace std;
using ll = long long;
template<class T> bool cmin(T &i, T j) { return i > j ? i=j,1:0; }
template<class T> bool cmax(T &i, T j) { return i < j ? i=j,1:0; }
const int N=1e5;
vector<pair<int,int>> adj[N];
int dp[N],dp2[N];
bool done[N];
vector<int> cc;
void getCC(int u,int e) {
    cc.push_back(u);
    for (auto &[v,len]: adj[u]) if (v!=e) {
        getCC(v,u);
    }
}
void dfs(int u,int e) {
    for (auto &[v,len]: adj[u]) if (v!=e) {
        dfs(v,u);
        cmax(dp[u],dp[v]+len);
    }
}
void dfs2(int u,int e) {
    multiset<int> lengths;
    lengths.insert(dp2[u]);
    for (auto &[v,len]: adj[u]) if (v!=e) {
        lengths.insert(dp[v]+len);
    }
    for (auto &[v,len]: adj[u]) if (v!=e) {
        lengths.erase(lengths.find(dp[v]+len));
        assert(!lengths.empty());
        dp2[v]=*lengths.rbegin()+len;
        dfs2(v,u);
        lengths.insert(dp[v]+len);
    }
}
int travelTime(int n, int m, int k, int u[], int v[], int len[]) {
    for (int i=0;i<m;i++) {
        adj[u[i]].emplace_back(v[i],len[i]);
        adj[v[i]].emplace_back(u[i],len[i]);
    }
    vector<int> diams;
    const int inf = 1e9;
    for (int i=0;i<n;i++) {
        if (!done[i]) {
            cc.clear();
            getCC(i,-1);
            for (int &j: cc) {
                assert(!done[j]);
                done[j]=true;
            }
            dfs(i,-1);
            dfs2(i,-1);
            int rs=inf;
            for (int &j: cc) {
                cmin(rs,max(dp[j],dp2[j]));
            }
            diams.push_back(rs);
            // dp[u] = max dist in subtree of u,
            // dp2[u] = max dist in everything except subtree of u
        }
    }
    sort(diams.begin(),diams.end());
    n = int(diams.size());
    if (diams.size()==1) {
        return diams[0];
    } else if (diams.size()==2) {
        return diams[0]+diams[1]+k;
    } else {
        return max(diams[n-1]+diams[n-2]+k,diams[n-2]+diams[n-3]+2*k);
    }
}

• Subtask 1: 0 / 14
• Subtask 2: 0 / 10
• Subtask 3: 0 / 23
• Subtask 4: 18 / 18
• Subtask 5: 0 / 12
• Subtask 6: 0 / 23