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_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


队列，首先给出一个顺序，先把数据读入队列，初始ran为1，然后开始扫队列，每次扫一遍，把每ng个中最大的加到队列尾部，其他的出队，同时更新每个number的ran,以及record。详见代码：

#include <iostream>
#include <queue>
#include <cstdio>
using namespace std;
struct ranks
{
    int num,w,r;///num是每个programmer对应的number  w是对应weight r是代表第几波被扫除队列的
    ranks(){}
    ranks(int numm,int ww,int rr):num(numm),w(ww),r(rr){}
};
int record[1000];///第i波被扫除队列的元素个数
int main()
{
    int np,ng;///np ng如题
    int s[1000],p,ran[1000],maxv;///s记录w ran记录对应number的r（即第几波被扫除）maxv记录最后一波只剩一个元素时的r
    queue<ranks> q;
    scanf("%d %d",&np,&ng);
    for(int i = 0;i < np;i ++)
    {
        scanf("%d",&s[i]);
    }
    for(int i = 0;i < np;i ++)
    {
        scanf("%d",&p);
        q.push(ranks(p,s[p],1));
    }
    record[1] = np;
    while(!q.empty())
    {
        int ma = 0,qsize = q.size();
        ranks wh,temp;
        for(int i = 1;i <= qsize;i ++)
        {
            temp = q.front();///队首给temp
            q.pop();///出队
            record[temp.r] ++;///对应更新record
            ran[temp.num] = temp.r;///更新ran
            if(temp.w > ma)ma = temp.w,wh = temp;///更新ma最大值
            if(i % ng == 0 || i == qsize)///每ng个元素一个group
            {
                record[wh.r] --;///不该在这一波扫除，应该在下一波 对应record 改变
                wh.r ++;
                q.push(wh);
                ran[wh.num] = wh.r;
                ma = 0;
            }
        }
        if(q.size() == 1)///只剩一个元素则不需要在比较了
        {
            record[wh.r] ++;///不能忘记更新record
            maxv = wh.r;
            break;
        }
    }
    for(int i = maxv - 1;i >= 1;i --)
    {
        record[i] += record[i + 1];///倒着来看，越晚被扫除排名越高，记录自己之后被扫除的人数，方便输出排名
    }
    for(int i = 0;i < np;i ++)
    {
        int rand = record[ran[i] + 1] + 1;
        if(i == 0)printf("%d",rand);
        else printf(" %d",rand);
    }
}

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