Dungeons

dungeons
English (ISC)
Dungeons Game
Robert is designing a new computer game. The game involves one hero, n opponents and n + 1
dungeons. The opponents are numbered from 0 to n −1 and the dungeons are numbered from 0
to n. Opponent i ( 0 ≤i ≤n −1) is located in dungeon i and has strength s[i]. There is no
opponent in dungeon n.
The hero starts off entering dungeon x, with strength z. Every time the hero enters any dungeon i (
0 ≤i ≤n −1), they confront opponent i, and one of the following occurs:
If the hero's strength is greater than or equal to the opponent's strength s[i], the hero wins. This
causes the hero's strength to increase by s[i] ( s[i] ≥1). In this case the hero enters dungeon
w[i] next ( w[i] > i).
Otherwise, the hero loses. This causes the hero's strength to increase by p[i] ( p[i] ≥1). In
this case the hero enters dungeon l[i] next.
Note p[i] may be less than, equal to, or greater than s[i]. Also, l[i] may be less than, equal to, or
greater than i. Regardless of the outcome of the confrontation, the opponent remains in dungeon i
and maintains strength s[i].
The game ends when the hero enters dungeon n. One can show that the game ends after a finite
number of confrontations, regardless of the hero's starting dungeon and strength.
Robert asked you to test his game by running q simulations. For each simulation, Robert defines a
starting dungeon x and starting strength z. Your task is to find out, for each simulation, the hero's
strength when the game ends.
Implementation details
You should implement the following procedures:
void init(int n, int[] s, int[] p, int[] w, int[] l)
n: number of opponents.
s, p, w, l: arrays of length n. For 0 ≤i ≤n −1:
s[i] is the strength of the opponent i. It is also the strength gained by the hero after
winning against opponent i.
p[i] is the strength gained by the hero after losing against opponent i.
w[i] is the dungeon the hero enters after winning against opponent i.
l[i] is the dungeon the hero enters after losing against opponent i.
This procedure is called exactly once, before any calls to simulate (see below).
Dungeons (1 of 4)

int64 simulate(int x, int z)
x: the dungeon the hero enters first.
z: the hero's starting strength.
This procedure should return the hero's strength when the game ends, assuming the hero starts
the game by entering dungeon x, having strength z.
The procedure is called exactly q times.
Example
Consider the following call:
init(3, [2, 6, 9], [3, 1, 2], [2, 2, 3], [1, 0, 1])
The diagram above illustrates this call. Each square shows a dungeon. For dungeons 0, 1 and 2,
the values s[i] and p[i] are indicated inside the squares. Magenta arrows indicate where the hero
moves after winning a confrontation, while black arrows indicate where the hero moves after losing.
Let's say the grader calls simulate(0, 1) .
The game proceeds as follows:
Dungeon
Hero's strength before confrontation
Result
0
1
Lose
1
4
Lose
0
5
Win
2
7
Lose
1
9
Win
2
15
Win
3
24
Game ends
As such, the procedure should return 24.
Dungeons (2 of 4)

Let's say the grader calls simulate(2, 3) .
The game proceeds as follows:
Dungeon
Hero's strength before confrontation
Result
2
3
Lose
1
5
Lose
0
6
Win
2
8
Lose
1
10
Win
2
16
Win
3
25
Game ends
As such, the procedure should return 25.
Constraints
1 ≤n ≤400 000
1 ≤q ≤50 000
1 ≤s[i], p[i] ≤10 (for all 0 ≤i ≤n −1)
0 ≤l[i], w[i] ≤n (for all 0 ≤i ≤n −1)
w[i] > i (for all 0 ≤i ≤n −1)
0 ≤x ≤n −1
1 ≤z ≤10
Subtasks
1. (11 points) n ≤50 000, q ≤100, s[i], p[i] ≤10 000 (for all 0 ≤i ≤n −1)
2. (26 points) s[i] = p[i] (for all 0 ≤i ≤n −1)
3. (13 points) n ≤50 000, all opponents have the same strength, in other words, s[i] = s[j] for
all 0 ≤i, j ≤n −1.
4. (12 points) n ≤50 000, there are at most 5 distinct values among all values of s[i].
5. (27 points) n ≤50 000
6. (11 points) No additional constraints.
Sample grader
The sample grader reads the input in the following format:
line 1: n q
line 2: s[0] s[1] ... s[n −1]
line 3: p[0] p[1] ... p[n −1]
7
7
Dungeons (3 of 4)

line 4: w[0] w[1] ... w[n −1]
line 5: l[0] l[1] ... l[n −1]
line 6 + i ( 0 ≤i ≤q −1): x z for the i-th call to simulate .
The sample grader prints your answers in the following format:
line 1 + i ( 0 ≤i ≤q −1) : the return value of the i-th call to simulate .
Dungeons (4 of 4)

Input / Output on this judge

This is the IOI function-implementation task dungeons adapted to standard input / output. Instead of implementing the function, read its arguments from standard input and print the returned value(s) to standard output, exactly as the official grader below does (its internal anti-cheat checks have been removed). You may also simply submit the grader together with your own implementation of the function.

Function signature (dungeons.h):

#include <vector>

void init(int n, std::vector<int> s, std::vector<int> p, std::vector<int> w, std::vector<int> l);
long long simulate(int x, int z);

Reference I/O driver (official grader, sanitized):

#include "dungeons.h"
#include <vector>
#include <cassert>
#include <cstdio>

static int n, q;
static std::vector<int> s, p, z;
static std::vector<int> w, l, x;
static std::vector<long long> answer;

int main() {
    assert(scanf("%d %d", &n, &q) == 2);
    s.resize(n);
    p.resize(n);
    w.resize(n);
    l.resize(n);
    x.resize(q);
    z.resize(q);
    answer.resize(q);

    for (int i = 0; i < n; i++) {
        assert(scanf("%d", &s[i]) == 1);
    }
    for (int i = 0; i < n; i++) {
        assert(scanf("%d", &p[i]) == 1);
    }
    for (int i = 0; i < n; i++) {
        assert(scanf("%d", &w[i]) == 1);
    }
    for (int i = 0; i < n; i++) {
        assert(scanf("%d", &l[i]) == 1);
    }


    init(n, s, p, w, l);

    for (int i = 0; i < q; i++) {
        assert(scanf("%d %d", &x[i], &z[i]) == 2);
        answer[i] = simulate(x[i], z[i]);
    }
    fclose(stdin);

    for (int i = 0; i < q; i++) {
        printf("%lld\n", answer[i]);
    }
    fclose(stdout);
    return 0;
}

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출처 IOI 2021