PAT (Advanced Level) 1056. Mice and Rice (25) 模拟比赛，用queue辅助

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

输入的第三行表示参赛者的序号。代码的框架跟层序遍历一样，用两个queue进行辅助，一个为当前层，一个为下一层。

/*2015.7.27cyq*/#include <iostream>#include <vector>#include <queue>#include <fstream>using namespace std;//ifstream fin("case1.txt");//#define cin finstruct mouse{int num;int weight;};int main(){int N,G;cin>>N>>G;vector<mouse> mice(N);vector<int> result(N);//最终排名for(int i=0;i<N;i++){mice[i].num=i;cin>>mice[i].weight;}queue<mouse> cur;int x;for(int i=0;i<N;i++){cin>>x;cur.push(mice[x]);}int rank=N/G+1;//比晋级者的排名多1if(N%G)//不足G只老鼠的的组还有晋级者rank++;queue<mouse> next;while(cur.size()!=1){while(!cur.empty()){//一组一组处理vector<mouse> group;int left=cur.size();for(int i=0;i<G&&i<left;i++){group.push_back(cur.front());cur.pop();}left-=G;int fat=-1;int winner=-1;for(int i=0;i<group.size();i++){if(group[i].weight>fat){fat=group[i].weight;winner=group[i].num;}}for(int i=0;i<group.size();i++){if(group[i].num==winner)next.push(group[i]);elseresult[group[i].num]=rank;}}//所有组都处理完后swap(cur,next);rank=cur.size()/G+1;if(cur.size()%G)rank++;}result[cur.front().num]=1;cout<<result[0];for(int i=1;i<N;i++)cout<<" "<<result[i];return 0;}