1056. Mice and Rice (25)

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:

11 325 18 0 46 37 3 19 22 57 56 106 0 8 7 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

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若干人随机分组进行比赛（可视为晋级赛），求每个人的排名。模拟这个晋级赛的过程。在求排名的函数中，对给出的排列，按每组m人（最后一组小于等于m人）分成若干组，对每一组组内进行比较，选出值最高的进入下一轮，用一个vector数组储存该人的序号。在对这一轮的全部人循环的同时，更新每个人的排名为当前的排名，而当前的排名就是进入下一轮的人的总数加一。然后以进入下一轮的人的序号数组为参数，递归调用求排名的函数。这样到最后就能得到正确的排名。

代码：

#include <iostream>#include <cstring>#include <vector>#include <cstdlib>#include <cstdio>using namespace std;vector<int>weight;vector<int>ranks;void func(vector<int>order,int m){int n=order.size();if(n==1) {ranks[order[0]]=1;return;}vector<int>next;int cur_rank=(n/m)+((n%m)==0?0:1)+1;for(int i=0;i<n;){int Max=-1,index,j;for(j=i;j<n&&j<i+m;j++){if(Max<weight[order[j]]){Max=weight[order[j]];index=order[j];}ranks[order[j]]=cur_rank;}next.push_back(index);i=j;}func(next,m);}int main(){int n,m;cin>>n>>m;weight.resize(n);for(int i=0;i<n;i++){cin>>weight[i];}vector<int>order(n);for(int i=0;i<n;i++){cin>>order[i];}ranks.resize(n);func(order,m);cout<<ranks[0];for(int i=1;i<n;i++) cout<<" "<<ranks[i];}