1056. Mice and Rice (25)-PAT

1056. Mice and Rice (25)

时间限制

30 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

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_Gwinners 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_iis 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

推荐指数：※

来源：http://pat.zju.edu.cn/contests/pat-a-practise/1056

题目很长，讲了一大段背景，核心就一句话，每一人一个体重，分组比较，赢得继续，直到冠军产生。（没做优化，竟然0ms通过。。，测试集比较小吧）

#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;}