浙大2013复试：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_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 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

题意如果读明白了，其实就是比赛的模拟，胜利者进入下一轮，失败者的排名相同，最后输出各个mice的最终排名。注意inital playing order代表的意思是比赛的次序，即前m个序号的mice在一起比，下一组m个序号的mice在一起比等等。采用队列，可以很方便的模拟出最终的排名。

代码如下：

#include<iostream>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;
int mice[1000];
int r[1000];
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("D:\\in.txt", "r", stdin);
    freopen("D:\\out.txt", "w", stdout);
#endif
    int n(0), m(0);
    cin >> n >> m;
    for (int i = 0; i < n; i++)
    {
        cin >> mice[i];
    }
    queue<int> order;
    int t(0);
    for (int i = 0; i < n; i++)
    {
        cin >> t;
        order.push(t);
    }
    while(order.size() != 1)
    {
        int gamecount = order.size() / m + (order.size() % m == 0 ? 0 : 1);
        int curank = gamecount + 1;//这一轮比赛失败者的排名
        queue<int> tmp;
        for (int i = 0; i < gamecount; i++)
        {
            int max = -1;
            int maxindex = -1;
            for (int j = i*m; j < i*m + m&&order.size()!=0; j++)//如果不满m个，需要判断order.size()!=0
            {
                int index = order.front();
                order.pop();
                if (mice[index]>max)
                {
                    max = mice[index];
                    if (maxindex != -1)
                    {
                        r[maxindex] = curank;
                    }
                    maxindex = index;
                }
                else
                {
                    r[index] = curank;
                }
            }
            tmp.push(maxindex);//一小组比赛结束后，把胜利者的序号放入队列
        }
        order = tmp;
    }
    r[order.front()] = 1;
    cout << r[0];
    for (int i = 1; i < n; i++)
    {
        cout << " " << r[i];
    }
    cout << endl;
    return 0;
}