Submission #1167493

#	Time	Username	Problem	Language	Result	Execution time	Memory
1167493		dragst	The Ties That Guide Us (CEOI23_incursion)	C++20	30 / 100	92 ms	6804 KiB

incursion

#include <bits/stdc++.h>
#include "incursion.h"
using namespace std;

int root, h[50005], p[50005], l[50005], r[50005], sz[50005];
vector<int> adj[50005];

void dfs(int x)
{
    sz[x]=1;
    for (auto y: adj[x])
    {
        if (y!=p[x] && l[x]==0)
        {
            p[y]=x; l[x]=y;
            h[y]=h[x]+1;
            dfs(y);
            sz[x]+=sz[y];
        }
        else if (y!=p[x])
        {
            p[y]=x; r[x]=y;
            h[y]=h[x]+1;
            dfs(y);
            sz[x]+=sz[y];
        };
    };
}

vector<int> mark(vector<pair<int, int>> F, int safe)
{
    int n=1, i;
    for (auto p: F)
    {
        n=max(n, p.first);
        n=max(n, p.second);
        adj[p.first].push_back(p.second);
        adj[p.second].push_back(p.first);
    };
    for (i=1; i<=n; i++)
    {
        if (adj[i].size()==2) {root=i; break;};
    };
    h[root]=1;
    dfs(root);
    vector<int> T(n);
    for (i=0; i<n; i++) {T[i]=0;};
    while (h[safe]>=1) {T[safe-1]=1; safe=p[safe];};
    return T;
}

void locate(vector<pair<int, int>> F, int curr, int t)
{
    int n=1, i;
    for (auto p: F)
    {
        n=max(n, p.first);
        n=max(n, p.second);
        adj[p.first].push_back(p.second);
        adj[p.second].push_back(p.first);
    };
    for (i=1; i<=n; i++)
    {
        if (adj[i].size()==2) {root=i; break;};
    };
    h[root]=1;
    dfs(root);
    if (t==0)
    {
        do {curr=p[curr]; i=visit(curr);}
        while (i==0);
    };
    if (l[curr]==0) {return;};
    do
    {
        if (sz[l[curr]]<=sz[r[curr]]) {curr=r[curr];}
        else {curr=l[curr];};
        i=visit(curr);
        if (i==0)
        {
            curr=p[curr];
            i=visit(curr);
            if (sz[l[curr]]<=sz[r[curr]]) {curr=l[curr];}
            else {curr=r[curr];};
            i=visit(curr);
        };
    }
    while (l[curr]>0 && i==1);
    if (i==0) {curr=p[curr]; i=visit(curr);};
    return;
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 30
• Subtask 2: 30 / 30
• Subtask 3: 0 / 40