PAT 1056. Mice and Rice (25)

1056. Mice and Rice (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

水题。

#include<string>#include<stdio.h>#include<iostream>#include<string.h>#include<queue>#include<algorithm>#include<map>#include<vector>using namespace std;int weigh[1005];int turn[1005];int wins[1005]={0}; //胜场int contest[1005];  //擂台里的人 int main(){   int np,ng;   cin>>np>>ng;   for(int i=0;i<np;i++)     scanf("%d",&weigh[i]);   for(int i=0;i<np;i++)     scanf("%d",&turn[i]);     int p=0;     int mem;   while(1)   {       int cnt=0;       int flag=0;      for(int i=0;i<np;i++)      {         if(wins[turn[i]]==p)         {             flag++;             mem=turn[i];           contest[cnt++]=turn[i];         }          if(cnt==ng||i==np-1)          {              if(cnt==0) continue;              int maxx=contest[0];              for(int j=0;j<cnt;j++)                if(weigh[contest[j]]>weigh[maxx]) maxx=contest[j];              wins[maxx]++;              cnt=0;          }      }        p++;        if(flag==1) break;   }       wins[mem]--;       int flaag=0;       int last[1005];       int zzz=0;       int mmm=1;         for(int i=wins[mem];i>=0;i--)        {            zzz+=mmm;            mmm=0;            for(int j=0;j<np;j++)            if(wins[j]==i)            {                last[j]=zzz;                 mmm++;            }        }         for(int i=0;i<np;i++)         {             if(flaag==1) printf(" ");             flaag=1;             printf("%d",last[i]);         }    return 0;}