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

思路：使用队列的方法让老鼠不断的入队出队，实现每轮次的晋级操作。

#include<stdio.h>#include<stdlib.h> #include<math.h>#include<string.h>#include<algorithm>#include <vector> #include<stack> #include <queue>  #include<map>#include<iostream>  #include <functional> #define MAX 1001#define MAXD 6#define TELNUM 10using namespace std;struct mouse{int weight;int rank;}mouse[MAX];int main(void){int Np,Ng;queue<int> q;scanf("%d%d",&Np,&Ng);for(int i = 0; i < Np; i++){scanf("%d",&mouse[i].weight);}int n;for(int i = 0; i < Np; i++){scanf("%d",&n);q.push(n);/*将老鼠的编号依次入队*/}int temp = Np;int group;while(q.size() != 1)/*按照队列中老鼠序号的排序，选出每轮晋级的老鼠，直到选出队列中最大的那只老鼠*/{if(temp % Ng == 0)/*把老鼠分别分成Ng个一组，最后不够Ng个单独为一组*/{group = temp / Ng;}else{group = temp / Ng + 1;}for(int i = 0; i < group; i++)/*选出小组中质量大的那个晋级*/ {int k = q.front();/*k表示质量最大的老鼠编号*/for(int j = 0; j < Ng; j++){if(i * Ng + j >= temp)break;int front = q.front();if(mouse[k].weight < mouse[front].weight){k = front;}mouse[front].rank = group + 1;/*该轮的老鼠名次都为group + 1*/q.pop();/*该老鼠出队*/}q.push(k);/*将小组最大的老鼠再次入队*/}temp = group;/*此轮有group只老鼠晋级，这group只老鼠是下轮比较的总数*/}mouse[q.front()].rank = 1;/*最后选出来的老鼠排名为 1*/for(int i = 0; i < Np; i++){printf("%d",mouse[i].rank);if(i < Np - 1)printf(" ");}return 0;}