1056. Mice and Rice (25)PAT

1. 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 NP programmers. Then every NG 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 NG 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: NP and NG (<= 1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than NG 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 NP distinct non-negative numbers Wi (i=0,…NP-1) where each Wi is the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,…NP-1 (assume that the programmers are numbered from 0 to NP-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：以题目给出的样例来说：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。
3、这里有一点比较关键的就是未晋级的mice的排名名次的计算，知道了当前要比赛的mice的数量之后就可以计算出一共会有多少晋级，那么未晋级的mice的名次就都是本轮晋级数目+1了。
因为每次都要筛选，所以要不停的更新数组。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
struct node
{
   int w,id,lev;
}p[1001];
int order[1001],order1[1001],rankk[1001];
int cmpp(const void *a,const void *b)
{
    return ((node *)b)->lev-((node *)a)->lev;
}
int main()
{
    int n,m,i;
    scanf("%d%d",&n,&m);
    for(i=0;i<n;i++)
    {
        scanf("%d",&p[i].w);
        p[i].id=i;
        p[i].lev=0;
    }
    for(i=0;i<n;i++)
        scanf("%d",&order[i]);
    int count,max,maxi,k=0;
    int tmp=n;
    while(1)
    {
       max=-1;k=0;count=0;
       for(i=0;i<tmp;i++)
       {
          if(p[order[i]].w>max)
          {
              maxi=i;
              max=p[order[i]].w;
          }
          count++;
          if(count==m||(i==tmp-1&&count<m))
          {
              max=-1;
              count=0;
              order1[k++]=order[maxi];///选出每一组的最大，然后初始化
          }
       }
       for(i=0;i<k;i++)
       {
           p[order1[i]].lev++;
           order[i] = order1[i];
       }
       tmp = k;///每次筛选出每一组的冠军
       if(k==1) break;
    }
    qsort(p,n,sizeof(p[0]),cmpp);///根据lev排序
    rankk[p[0].id]=1;
    for(i=1;i<n;i++)
    {
        if(p[i].lev==p[i-1].lev) rankk[p[i].id]=rankk[p[i-1].id];
        else rankk[p[i].id]=i+1;
    }
    printf("%d",rankk[0]);
    for(i=1;i<n;i++)
        printf(" %d",rankk[i]);
    printf("\n");
    return 0;
}