【PAT】1056. Mice and Rice (25)

 

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_Gprogrammers 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_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 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

这道题目的题意不好懂，博客： http://blog.youkuaiyun.com/yangsongtao1991/article/details/43417363   对题目的分析比较具体，如下：

     1）input中的第三行是索引，表示的是以第三行索引的顺序进行比较，也就是说第6、0、8个mice先组成一组进行比较，而这3个mice吃到的rice就是第二行中对应的值19、25、57，因此第8个mice吃到的是57数量最多晋级，同理第7、10、5个mice吃到的rice是22、10、3，第7个晋级，第9、1、4个mice吃到的rice是56、18、37，第9个晋级，剩下的只有第2、3个mice，分别吃到的rice是0、46，第3个晋级，于是第8、7、9、3一共4个mice晋级到下一轮，并继续按照这个顺序比较，那么其他没有晋级的mice的排名都是第5名。
     2）第8、7、9、3个mice吃到的rice分别是57、22、56、46，前三个中第8个晋级，剩下的第3个也晋级，所以未晋级的7、9的排名都是第3。
     3）第8、3个mice的排名分别是第1和第2了。

知道题意后，就好做了，就是模拟。

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

struct mice{
	mice(){
	}
	mice(int w, int r, int i){
		wei = w;
		rank = r;
		index = i;		
	}
	int wei;
	int rank;
	int index;	
};

void cmp(vector<mice> &mices,int loserRank,vector<int> t, vector<int> &winners){
	int i, max, winnerIndex, wei;
	winnerIndex = t[0];
	max = mices[t[0]].wei;
	for(i=1; i<t.size(); i++){
		wei = mices[t[i]].wei;
		if(wei > max){
			winnerIndex = t[i];
			max = wei;
		}
	}
	for(i=0; i<t.size(); i++){
		if(t[i]!=winnerIndex){
			mices[t[i]].rank = loserRank;
		}
	}	
	winners.push_back(winnerIndex);
}

int main(int argc, char** argv) {
	int np, ng, i, wei;
	cin>>np>>ng;
	vector<mice> mices;
	vector<mice> tmp(np);
	for(i=0; i<np; i++){
		cin>>wei;
		mices.push_back( mice(wei,0,i) );
		tmp.push_back( mice(wei,0,i) );
	}
	int cnt=0, pos, loserRank;
	vector<int> t;
	vector<int> vec;
	vector<int> winners;
	for(i=0; i<np; i++){
		cin>>pos;
		vec.push_back(pos);
	}	
	
	while(vec.size() != 1){ 	
		if(vec.size()%ng == 0){			
			loserRank = vec.size()/ng + 1;
		}else{
			loserRank = vec.size()/ng + 2;		
		}	
		for(i=0; i<vec.size(); i++){
			t.push_back(vec[i]);
			cnt++;
			if(cnt==ng || i==vec.size()-1){
				cmp(mices, loserRank, t, winners);
				t.clear();
				cnt = 0;
			}		
		}	 
		vec.assign(winners.begin(), winners.end());
		winners.clear();
	}
	mices[vec[0]].rank = 1;
	cout<<mices[0].rank;
	for(i=1; i<mices.size(); i++){
		cout<<" "<<mices[i].rank;
	}	
	cout<<endl;	
	return 0;
}