1056 Mice and Rice (25分) 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 3Sample Output:
5 5 5 2 5 5 5 3 1 3 5 #include<bits/stdc++.h> using namespace std; const int maxn=1010; int w[maxn],ans[maxn]; int main(){ int n,m; cin>>n>>m; vector<int>cur(n); for(int i=0;i<n;i++)cin>>w[i]; for(int i=0;i<n;i++)cin>>cur[i]; while(cur.size()>1){ vector<int>next; int remain=(cur.size()+m-1)/m; for(int i=0;i<cur.size();){ int j=min(i+m,(int)cur.size()); int t=i; for(int k=i;k<j;k++){ if(w[cur[k]]>w[cur[t]]) t=k; } next.push_back(cur[t]); for(int k=i;k<j;k++){ if(k!=t) ans[cur[k]]=remain+1; } i=j; } cur=next; } ans[cur[0]]=1; cout<<ans[0]; for(int i=1;i<n;i++)cout<<' '<<ans[i]; return 0; }