P7167 [eJOI2020 Day1] Fountain 题解

题目链接：Fountain

很不错的基础算法组合题：单调栈+倍增

1. 首先考虑到一个事实，就是下面第一个比当前半径大的位置会成为移动的第一次落脚点，抽象下就是下面出现的第一次比自身大的半径，这个问题显然可以单调栈轻松解决。

2. 第二个点就是我们知道了单次移动的第一个位置，现在问你多次移动到的位置，并且还要保证容量是恰好大于上一次移动的喷泉圆盘，小于当前喷泉的圆盘。这点显然很容易用倍增处理出来，这若干个圆盘跳跃多次的落脚点以及此时此刻的总容量。

3. 最后的细节自然是注意到 \(0\) 处的容量其实是无穷大，而 \(0\) 处对应的恰好又是跳出喷泉的地方，所以初始化容量为无穷大，这样一来保证不在喷泉内的地方并不影响倍增答案，当恰好大于总的容量，倍增则恰好跳到了 \(0\) 处。

参照代码

#include <bits/stdc++.h>

// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC optimize(2)

#define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}


template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

constexpr int N = 1e5 + 10;
constexpr int T = 25;
ll sum[N][T], go[N][T];
stack<int> st;
pii node[N];
int n, q;

inline void solve()
{
    cin >> n >> q;
    const int stepMax = log2(n);//倍增上限步长
    forn(i, 1, n)forn(j, 0, stepMax)sum[i][j] = 1e9;//初始化为所有地方无穷大
    forn(i, 1, n)cin >> node[i].first >> node[i].second, sum[i][0] = node[i].second;//初始化要跳出第一步需要的容量
    forn(i, 1, n)
    {
        while (!st.empty() and node[st.top()].first < node[i].first)
        {
            go[st.top()][0] = i;//更新第一次条跳的位置
            st.pop();
        }
        st.push(i);
    }
    //倍增预处理
    forn(j, 1, stepMax)
    {
        forn(i, 1, n-(1<<j)+1)
        {
            go[i][j] = go[go[i][j - 1]][j - 1];
            sum[i][j] = sum[i][j - 1] + sum[go[i][j - 1]][j - 1];
        }
    }
    while (q--)
    {
        ll pos, val;
        cin >> pos >> val;
        forv(i, stepMax, 0)if (val > sum[pos][i])val -= sum[pos][i], pos = go[pos][i];//能大于这个容量就跳
        cout << pos << endl;
    }
}

signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

\[时间复杂度为 \ O((N+Q)\log{N}) \]