Problem
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.
（第一段相当于故事背景，没啥用。。）
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.
（第二、三段来说明一下游戏规则）意思就是一共有NP个老鼠，每NG个一组，最后不够NG个的，单独分一组，每一轮中每组胜利的那一个进入下一轮，该轮失败的所有老鼠，他们的排名是相同的，一直这样做，直到剩下最后一个。排名第一，依次类推。
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.
为了使问题简单化，假设每个老鼠的重量是不变的，输入是老鼠的重量以及老鼠的排列序号，需要输出最终每个老鼠的排名。
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers: NP and NG (≤1000), 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 (i=0,⋯,NP−1) 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,⋯,NP−1 (assume that the programmers are numbered from 0 to NP−1). All the numbers in a line are separated by a space.
每一个输入包括三行，第一行是NP和NG的值，分别代表的是老鼠的个数以及每个组的大小。第二行中每个老鼠的重量是互不相同的，第三行是每个老鼠的排列序号，或者说是出场顺序。（说不清楚，请看我下面的例子解析。）
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.
常规输出要求。不细说了。
Sample Input:
11 3
25 18 0 46 37 3 19 22 57 56 10 这一行代表的是每个老鼠的重量。
6 0 8 7 10 5 9 1 4 2 3 //这一行第一个数代表的是第6号老鼠第一出场。0号老鼠第二出场。
Sample Output:
5 5 5 2 5 5 5 3 1 3 5
我的思路
定义一个map表示一个老鼠的重量和输出序号，默认都初始化为1。
定义一个队列，按照第三行的顺序把对应的元素push进去。
然后，每NG个一组，找出最大值，最大值，加入队列，普通值加入一个栈，每次一轮结束之后，输出栈的元素，并赋值为当前队列的大小加1。
（大致是这样，具体细节处理，见下面代码。）
My Solution
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <time.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstring>
#include <map>
#include <iterator>
using namespace std;
#pragma warning(disable:4996)
int main()
{
map<int, int>map1;
int m = 0;
int n = 0;
cin >> m >> n;
int a[1000] = { 0 };
for (int i = 0; i < m; i++) {
cin >> a[i];
}
//首先初始化每个老鼠的输出序号是1
for (int i = 0; i < m; i++) {
map1[a[i]] = 1;
}
queue<int>d;
int c = 0;
for (int i = 0; i < m; i++) {
cin >> c;
d.push(a[c]);
}
int b[1000] = { 0 };
int i1 = 0;
stack<int>d2;
while (d.size()!=1) {
int j = d.size();
for (int j1 = 0; j1 < j; j1=j1+n) {
for (i1 = 0; i1 < n; i1++) {
if (j1+i1<j) {
b[i1] = d.front();
d.pop();
}
else break;
}
int max = -1;
//找到那个最大值。
for (int i = 0; i < i1; i++) {
if (b[i] > max) { max = b[i]; }
}
for (int i = 0; i < i1; i++) {
if (b[i] != max) { d2.push(b[i]); }
else { d.push(b[i]); }
}
}
int j3 = d.size();
while (!d2.empty()) {
map1[d2.top()] = j3 + 1;
d2.pop();
}
}
cout << map1[a[0]];
for (int i = 1; i < m; i++) {
cout << " " << map1[a[i]];
}
return 0;
}
另外，我看了算法笔记的解法，他使用结构体表示一个老鼠，其中一个元素是排名，就没有像我这样使用map，另外，他还注意到每轮的组数+1就是该轮被淘汰的老鼠的排名，这也是很仔细，优化了我代码中计算排名的那部分。其他的意思差不多。如果需要算法笔记的源码，可以CSDN私信我，或者留言。谢谢阅读！
