A1056. Mice and Rice (25)

1056. Mice and Rice (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

模拟排序

#include<cstdio>const int maxn = 1e3 + 10;int order[maxn], Rank[maxn], wgh[maxn], contest[maxn];  int main(){int np, ng;scanf("%d %d", &np, &ng);for(int i = 0; i < np; ++i)scanf("%d", &wgh[i]);for(int i = 0; i < np; ++i)  //编号和竞争顺序往后挪一位scanf("%d", &order[i]);while(1){int total = 0;for(int i = 0; i < np; ++i){if(!Rank[order[i]]){   //位序i还没有排序  ,则加入比较数组contest contest[total++] = order[i];   //total始终指向元素的后一位置}}if(total == 1){         //剩余最后一位胜者 Rank[contest[total - 1]] = 1;break;}int low = 0, high = 0;   //处理竞争区间 while(low < total){if(low % ng == 0){if(low + ng < total){high = low + ng;}else{high = total;}}else{++low;continue;}int now = low, MAX = -1;for(;now < high; ++now){if(wgh[contest[now]] > MAX){MAX = wgh[contest[now]];}}for(now = low; now < high; ++now){if(wgh[contest[now]] != MAX){Rank[contest[now]] = total / ng + 1 + (total % ng ? 1 : 0);}}++low;}}for(int i = 0; i < np; ++i)printf("%s%d", i ? " " : "", Rank[i]);return 0;}