Submission #1079210

#	Time	Username	Problem	Language	Result	Execution time	Memory
1079210		anango	Stations (IOI20_stations)	C++17	10 / 100	2177 ms	5172 KiB

stations

#include "stations.h"
#include <vector>
#include <bits/stdc++.h>
using namespace std;
#define cout cerr


vector<vector<int>> adjlist;

vector<int> depth;
vector<int> tin;
vector<int> tout;
int tim = 0;

void dfs(int node, int par) {
    tin[node] = tim++;
    for (int j:adjlist[node]) {
        if (j!=par) {
            depth[j] = depth[node]+1;
            dfs(j,node);
        }
    }
    tout[node] = tim++;
}

std::vector<int> label(int n, int k, std::vector<int> u, std::vector<int> v) {
	std::vector<int> labels(n);
    adjlist=vector<vector<int>>(n);
    tin=tout=vector<int>(n,-1);
    depth=vector<int>(n,0);
    for (int i=0; i<n-1; i++) {
        adjlist[u[i]].push_back(v[i]);
        adjlist[v[i]].push_back(u[i]);
    }
    dfs(0,-1);
	for (int i = 0; i < n; i++) {
        if (depth[i]%2==1) {
		    labels[i] = tin[i]+1;
        }
        else {
            labels[i] = 1000+tout[i]+1;
        }
        cout << i << " " << tin[i] <<" " << tout[i] <<" " << labels[i] << endl;
	}

    //consider using the euler tour trick on the tree
    //for each node, store the left and right times of its subtree in the label (using perfect hashing)
    //for a node, you know the tin and tout (call these inv and outv)
    //then for every child, you also know the tin and tout of that child (call these inc and outc)
    //and the tin and tout of the destination (call these ind and outd)
    //so you know that if inv<=ind<=outd<=outv, then the destination is in this subtree
    //and if inc<=ind<=outd<=outc for some child, the destination is in the subtree of this child
    //if none of these hold, then go up
    //this uses at most n^2 labels
    //we could just use in but then it jambloats because we don't know where the subtree ends
    //actually what about storing out for even depth
    //because we know one of the in values for child for free
    //and if we know out[node] we know one out value for free
    //that should work for 2000

	return labels;
}

int find_next_station(int s, int t, std::vector<int> c) {
    sort(c.begin(), c.end());
    tin[s];
    cout << "TRY FIND " << endl;
    for (auto i:c) {
        cout << i <<" ";
    }
    cout << endl;
    cout << "DOING " << s <<" " << t << endl;
    if (t>1000) {
        t-=1000;
    }
    if (s>1000) {
        //depth is even
        s-=1000;
        //so all the children are in-values, our node is an out-value
        //the lowest in-value is actually the parent, so that can be ignored
        c.push_back(s+1);
        for (int i=2; i<c.size(); i++) {
            if (c[i-1]<=t && t<c[i]) {
                return c[i-1];
            }
        }
        return c[0];
    }
    else {
        //depth is odd
        for (int i=0; i<c.size(); i++) {
            c[i]-=1000;
        }
        //so all the children are out-values, our node is an in-value
        //max value is actually the parent
        c.push_back(s+1);
        sort(c.begin(), c.end());
        cout << "DOCTORED C " << endl;
        for (auto i:c) {
            cout << i <<" ";
        }
        cout << endl;
        for (int i=1; i<((int)c.size())-1; i++) {
            if (c[i-1]<=t && t<=c[i]) {
                return c[i]+1000;
            }
        }
        return c.back()+1000;
    }
	return c[0];
}

Compilation message (stderr)

stations.cpp: In function 'int find_next_station(int, int, std::vector<int>)':
stations.cpp:82:24: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   82 |         for (int i=2; i<c.size(); i++) {
      |                       ~^~~~~~~~~
stations.cpp:91:24: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   91 |         for (int i=0; i<c.size(); i++) {
      |                       ~^~~~~~~~~

• Subtask 1: 0 / 5
• Subtask 2: 0 / 8
• Subtask 3: 0 / 16
• Subtask 4: 10 / 10
• Subtask 5: 0 / 61