제출 #770761

#제출 시각아이디문제언어결과실행 시간메모리
770761adrilen열대 식물원 (Tropical Garden) (IOI11_garden)C++17
69 / 100
5052 ms87800 KiB
#include "garden.h"
#include "gardenlib.h"
#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using arr = array<int, 2>;
using arrr = array<int, 3>;

constexpr int maxn = 1.5e5, maxg = 1e9, maxq = 2e3 + 5, bit = 32 - __builtin_clz(maxg), siz = 1 << bit;
int goal;

basic_string <int> adj[maxn];

// Jumping (1 << i) steps from each node, taking the best way or the second
// store whether we need to use the best or second best route
arr jump[bit][maxn][2] = { 0 };




void count_routes(int n, int m, int P, int R[][2], int Q, int G[])
{
    goal = P;

    for (int i = 0; i < m; i++)
    {
        adj[R[i][0]].push_back(R[i][1]);
        adj[R[i][1]].push_back(R[i][0]);
    }

    // Making the only way a second way if needed
    for (int i = 0; i < n; i++)
    {
        if (adj[i].size() == 1) adj[i].push_back(adj[i][0]);
    }


    for (int i = 0; i < n; i++)
    {
        for (int y = 0; y < 2; y++)
        {
            jump[0][i][y] = {adj[i][y], adj[adj[i][y]][0] == i};
        }
    }

    for (int b = 1; b < bit; b++)
    {
        for (int i = 0; i < n; i++)
        {
            for (int y = 0; y < 2; y++)
            {
                jump[b][i][y] = jump[b - 1][jump[b - 1][i][y][0]][jump[b - 1][i][y][1]];
            }
        }
    }

    vector <arr> queries(Q);

    for (int i = 0; i < Q; i++) queries[i] = { G[i], i };
    sort(queries.begin(), queries.end());

    arr pos;



    map<arr, int> pos_count, next_count;
    for (int i = 0; i < n; i++) pos_count[{i, 0}] = 1;

    int last = 0;
    int diff;


    vector <int> output(Q);


    for (int y = 0; y < Q; y++)     // 2e3
    {
        diff = queries[y][0] - last;

        for (auto pa : pos_count)
        {
            pos = pa.first;
            for (int j = 0; j < bit; j++) // 30
            {
                if (diff & (1 << j)) pos = jump[j][pos[0]][pos[1]];
            }
            next_count[pos] += pa.second;
        }
        output[queries[y][1]] = next_count[{P, 0}] + next_count[{P, 1}];
        last = queries[y][0];

        pos_count.swap(next_count);
        next_count.clear();
    }


    for (int i: output) answer(i);
}


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