1056 Mice and Rice

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​​ (≤), 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​​ (,) 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 (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

题意：

　　给出一个数组代表老鼠的重量，然后将这些老鼠按照给出的顺序，每Ng各一组，在每组中选出一个体重最重的，参与下一轮的比较，其余的老鼠淘汰且排名相同。输出每一个老鼠按照原来顺序的排名。

思路：

　　这道题考查的是queue的应用，做的时候应该理清楚里面的关系，然后在写代码。我们可以用每一轮选拔晋级的选手的个数，来确定这轮选拔中没有晋级的选手的排名，即没有晋级的选手的排名 = 晋级的选手的个数 + 1. 我们用一个struct来存储每一个老鼠的信息，主要包括：体重，本来的顺序和参加比赛时的出场顺序。当队列中只有一个元素的时候代表冠军已经产生。

Code：

#include <bits/stdc++.h>

using namespace std;

struct Mouse {
    int weight;
    int index0;
    int index;
    int rank;
};

bool cmp(Mouse a, Mouse b) { return a.index0 < b.index0; }

int main() {
    int Np, Ng, temp;
    cin >> Np >> Ng;
    vector<int> weight(Np);
    vector<Mouse> mouses(Np);
    queue<Mouse> que;
    for (int i = 0; i < Np; ++i) cin >> weight[i];
    for (int i = 0; i < Np; ++i) {
        cin >> temp;
        mouses[i] = {weight[temp], temp, i, 0};
        que.push(mouses[i]);
    }
    while (!que.empty()) {
        int size = que.size();
        if (size == 1) {
            Mouse tempMouse = que.front();
            mouses[tempMouse.index].rank = 1;
            break;
        }
        int group = size / Ng;
        if (size % Ng != 0) group += 1;
        int cnt = 0, maxn = -1;
        Mouse maxWeightMouse;
        for (int i = 0; i < size; ++i) {
            Mouse tempMouse = que.front();
            mouses[tempMouse.index].rank = group + 1;
            que.pop();
            cnt++;
            if (tempMouse.weight > maxn) {
                maxn = tempMouse.weight;
                maxWeightMouse = tempMouse;
            }
            if (cnt == Ng || i == size - 1) {
                cnt = 0;
                maxn = -1;
                que.push(maxWeightMouse);
            }
        }
    }
    sort(mouses.begin(), mouses.end(), cmp);
    for (int i = 0; i < Np; ++i) {
        if (i == 0)
            cout << mouses[i].rank;
        else
            cout << " " << mouses[i].rank;
    }
    return 0;
}