1056. Mice and rice (25)-pat

1056. Mice and rice (25) Time Limit 30 ms memory limit 32000 kb code length limit 16000 B discriminant program standard

Mice and riceIs 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 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 nGwinners 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 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 WIIS the weight of the I-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,... nThe P-1 (assume that the programmers are numbered from 0
To NThe 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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Source: http://pat.zju.edu.cn/contests/pat-a-practise/1056

The topic is very long. I spoke a lot about the background. The core is a single sentence. Each person has a weight, a group comparison, and won the championship until it is produced. (If NO optimization is performed, the test set is relatively small)

#include<iostream>#include<string.h>using namespace std;#define  N 1001int np,ng,w[N],permu[N],frank[N];void prank(int now_rank){int i,j;int sum_win;int fattest=-1,fat_num=-1;sum_win=0;j=0;for(i=0;i<np;i++){int now_pre=permu[i];if(j<ng){if(frank[now_pre]>=now_rank){j++;if(w[now_pre]>fattest){fattest=w[now_pre];fat_num=now_pre;}}}if((j==ng||i+1==np)&&fattest!=-1){frank[fat_num]++;sum_win++;fattest=-1;fattest=-1,fat_num=-1;j=0;}}if(sum_win>1){prank(now_rank+1);}}int putrank(int pr){int i,i_rank=1;for(i=0;i<np;i++)if(pr!=i&&frank[i]>frank[pr])i_rank++;return i_rank;}int main(){int i;cin>>np>>ng;for(i=0;i<np;i++)cin>>w[i];for(i=0;i<np;i++)cin>>permu[i];i=0;memset(frank,0,sizeof(frank));prank(0);cout<<putrank(0);for(i=1;i<np;i++)cout<<" "<<putrank(i);return 0;}