PAT (Advanced Level) Practice 1056 Mice and Rice (25 points) sorting, ranking]

Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse.

First the playing order is randomly decided for $N_P$ programmers. Then every $N_G$ programmers are grouped in a match. The fattest mouse in a group wins and enters the next turn. All the losers in this turn are ranked the same. Every $N_G$ winners are then grouped in the next match until a final winner is determined.

For the sake of simplicity, assume that the weight of each mouse is fixed once the programmer submits his/her code. Given the weights of all the mice and the initial playing order, you are supposed to output the ranks for the programmers.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: $N_P$ and $N_G (≤1000)$, the number of programmers and the maximum number of mice in a group, respectively. If there are less than $N_G$ mice at the end of the player’s list, then all the mice left will be put into the last group. The second line contains $N_P$ distinct non-negative numbers $W_i (i=0,⋯,N_P−1)$ where each $W_i$ is the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of $0,⋯,N_P−1$ (assume that the programmers are numbered from $0$ to $N_P−1$). All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the final ranks in a line. The $i$-th number is the rank of the $i$-th programmer, and all the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

11 3
25 18 0 46 37 3 19 22 57 56 10
6 0 8 7 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

The meaning of problems

Np is given mice, the mice will be ranked according to weight. More complex rules Ranking: these mice have an initial order, in that order of these rats were divided into groups of size Ng, pick the maximum weight of the mice in each group for the next round of comparison, other mice at the back and rankings all the same.

Thinking

This question recursive nature, could then write code.

Code

#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;

struct mouse {
    int rnk = -1;
    int weight = 0;
};

void set_rank(vector<mouse> &mice, vector<int> seq, int ng) {
    if (seq.size() == 1) { // 剩下一只老鼠，排名为1
        mice[seq[0]].rnk = 1;
        return;
    }

    int max_idx = seq[0]; // 重量最大的老鼠的序号
    vector<int> next_seq; // 下一轮比较的顺序
    for (int i = 1, j = 1, sz = seq.size(); i < sz; ++i, j = (j + 1) % ng) {
        if (mice[max_idx].weight < mice[seq[i]].weight)
            max_idx = seq[i];
        if (j == ng - 1 || i == sz - 1) {
            next_seq.push_back(max_idx);
            if (i + 1 < sz)
                max_idx = seq[i + 1];
        }
    }

    set_rank(mice, next_seq, ng); // 进行下一轮比较

    int rank = next_seq.size() + 1; // 本轮被淘汰的老鼠的排名
    for (int i = 0, sz = seq.size(); i < sz; ++i)
        if (mice[seq[i]].rnk == -1)
            mice[seq[i]].rnk = rank;
}

int main() {
    int np, ng;
    cin >> np >> ng;
    vector<mouse> mice(np);
    for (int i = 0; i < np; ++i)
        cin >> mice[i].weight;
    vector<int> seq(np);
    for (int i = 0; i < np; ++i)
        cin >> seq[i];

    set_rank(mice, seq, ng);

    for (int i = 0, sz = mice.size(); i < sz; ++i) {
        cout << mice[i].rnk;
        if (i != sz - 1)
            cout << " ";
    }
}