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 NP programmers. Then every NG 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 NG 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: NP and NG (≤), the number of programmers and the maximum number of mice in a group, respectively. If there are less than NG 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 NP distinct non-negative numbers Wi (,) where each Wi 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 NP−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:
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 struct Mouse { 6 int weight; 7 int index0; 8 int index; 9 int rank; 10 }; 11 12 bool cmp(Mouse a, Mouse b) { return a.index0 < b.index0; } 13 14 int main() { 15 int Np, Ng, temp; 16 cin >> Np >> Ng; 17 vector<int> weight(Np); 18 vector<Mouse> mouses(Np); 19 queue<Mouse> que; 20 for (int i = 0; i < Np; ++i) cin >> weight[i]; 21 for (int i = 0; i < Np; ++i) { 22 cin >> temp; 23 mouses[i] = {weight[temp], temp, i, 0}; 24 que.push(mouses[i]); 25 } 26 while (!que.empty()) { 27 int size = que.size(); 28 if (size == 1) { 29 Mouse tempMouse = que.front(); 30 mouses[tempMouse.index].rank = 1; 31 break; 32 } 33 int group = size / Ng; 34 if (size % Ng != 0) group += 1; 35 int cnt = 0, maxn = -1; 36 Mouse maxWeightMouse; 37 for (int i = 0; i < size; ++i) { 38 Mouse tempMouse = que.front(); 39 mouses[tempMouse.index].rank = group + 1; 40 que.pop(); 41 cnt++; 42 if (tempMouse.weight > maxn) { 43 maxn = tempMouse.weight; 44 maxWeightMouse = tempMouse; 45 } 46 if (cnt == Ng || i == size - 1) { 47 cnt = 0; 48 maxn = -1; 49 que.push(maxWeightMouse); 50 } 51 } 52 } 53 sort(mouses.begin(), mouses.end(), cmp); 54 for (int i = 0; i < Np; ++i) { 55 if (i == 0) 56 cout << mouses[i].rank; 57 else 58 cout << " " << mouses[i].rank; 59 } 60 return 0; 61 }