1056. Mice and Rice (25)

时间限制

30 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_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:

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

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这题做好好久，大受打击。第一题目比较难懂，最后写出了和题意不符。第二则是没有想到用队列，而是用了递归，结果十分麻烦。第三没有想到巧妙的排名方法。最后放弃了调试自己的代码，而是借鉴了网上的思路。

这里学习了queue 的使用方法。

#include <iostream>
#include <vector>
#include <queue>  
using namespace std;

int main()
{
	int Np,Ng;
	cin>>Np>>Ng;
	vector <int> weigh(Np);
	for (int i = 0; i < Np; i++)  
	{
		cin>>weigh[i];  
	}
	queue<int> order;
	int x;  
    for (int i = 0; i < Np; i++){  
        cin>>x; 
        order.push(x);  
    } 
	vector<int> rank(Np);
	while (order.size() != 1)
	{  
        queue<int> tmp;  
        int groupNum = order.size() / Ng + (order.size()%Ng == 0 ? 0:1); 
        for (int i = 0; i < groupNum; i++){  
            int max = -1, maxIndex = -1;  
            for (int j = i*Ng; j < i*Ng+ Ng && order.size()!= 0; j++)
			{  
                int t = order.front();  
                rank[t] = groupNum + 1;   //有几组，就代表有几人进入下一局，所以排名为下一组的人数加一
                order.pop();  
                if (weigh[t] > max)
				{  
                    max = weigh[t];  
                    maxIndex = t;  
                }  
            }  
            tmp.push(maxIndex);  
        }  
        order = tmp;  
    }
	rank[order.front()] = 1;  
	int  i;
	for (i = 0; i < Np-1; i++)  
        cout<<rank[i]<<" ";
	cout<<rank[i];
	return 0;
}