1056 Mice and Rice (25分)队列

1.27刷题2

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?? (≤), 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?? (,) 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 (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 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

// 25 18 0 46 37 3 19 22 57 56 10
// 6  0  8  7 10 5  9  1  4  2  3

//题没看懂，其次要不是这个小结是队列，我根本想不起来用这个 
#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
queue<int> q;
// int maxabc(int a,int b,int c){  //题目给的ng是3，我就鬼迷心窍写个这。。 
//     int max=-1;
//     max=a>b?a:b;
//     max=max>c?max:c;
//     return max;
// }
struct mouse {//不用结构体多麻烦呀，为啥不用呀 
    int weight,record;
} mice[maxn];
int main() {
    int np,ng,b;
    cin>>np>>ng;
    
    for(int i=0; i<np; i++)cin>>mice[i].weight;
    for(int i=0; i<np; i++) {
        cin>>b;
        q.push(b);
    }
    int temp=np,group;//没想到
    while(q.size()!=1) {//这个循环是真的大 ，没想到是分组来排名的，确实还是模拟，就是按照思路写的代码 
        if(temp%ng==0)group=temp/ng;
        else group=temp/ng+1;
        for(int i=0; i<group; i++) {
            int k=q.front();
            for(int j=0; j<ng; j++) {
                if(i*ng+j>=temp)break;
                int front=q.front();
                if(mice[front].weight>mice[k].weight)k=front;

                mice[front].record=group+1;//该轮老鼠排名为group+1
                q.pop();
            }
            q.push(k);//优胜者进决赛圈 
        }
        temp=group;//group在缩小 
    }
    mice[q.front()].record=1; 
    for(int i=0; i<np; i++) {
        cout<<mice[i].record;
        if(i<np-1)cout<<" ";
    }
}

明天还有实习，还有复试，还有毕设，还有过年，还有春晚，还有黑眼圈，还有卷。

1056 Mice and Rice (25分)队列

原文：https://www.cnblogs.com/xx123/p/14337635.html