1056 Mice and Rice （25 分)

2019-03-29

1056 Mice and Rice （25 分)

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 (≤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**i is 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:

1
2
3

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:

1	5 5 5 2 5 5 5 3 1 3 5

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

这题题目比较难懂，大意如下：

给出np老鼠总数，ng每组老鼠个数。需要按照ng分组，选出最重的老鼠进入下一轮。直到冠军产生为止。

第一行给出老鼠的个数以及每组老鼠数。第二行给出各老鼠重量。第三行给出老鼠的ID。以题中数据为例，ng为3，那么6，0，8为一组，选出最重的。7，10，5为一组，同样选出最重的，依次类推。每组中最重的进入下一轮，其余淘汰。输出各老鼠的排名。在同一轮中被淘汰的老鼠排名相同。

分析

利用四个vector<int>。winner，loser存放晋级和淘汰的老鼠id。temp存放当前所有老鼠，group存放一组老鼠。当winner不唯一时循环，每一轮的winner为下一轮考虑的temp。将temp中的老鼠按ng分组，选出最重的加入winner，其余加入loser。这样，一轮temp结束后，loser中老鼠的名次即为winner中老鼠数+1。如此循环，最后一只老鼠的排名即为1。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include<sstream>

using namespace std;
int weight[1010];
int rankk[1010];
int main() {

    int ng,np;
    cin>>np>>ng;
    int id;
    for(int i=0;i<np;i++)
        cin>>weight[i];
    vector<int> loser,winner,temp,group;
    for(int i=0;i<np;i++) {
        cin >> id;
        winner.push_back(id);
    }

    while (winner.size()!=1){
        temp=winner;
        winner.clear();
        loser.clear();
        group.clear();
        for(int i=0;i<temp.size();){
            int k=0,max=-1,maxid=-1;
            //按ng处理每组老鼠
            while(k<ng&&(i+k)<temp.size()){
                id=temp[i+k];
                group.push_back(id);
                //选出最重的
                if(weight[id]>max) {
                    max=weight[id];
                    maxid=id;
                }
                k++;
            }
            i+=k;
            winner.push_back(maxid);
            for(int i:group)
                if(i!=maxid)
                    loser.push_back(i);

        }
        //loser老鼠排名可确定
        for(int i:loser)
            rankk[i]=winner.size()+1;
    }
    rankk[winner[0]]=1;
    cout<<rankk[0];
    for(int i=1;i<np;i++)
        cout<<" "<<rankk[i];
    return 0;
}