PAT(甲级)1056

1056. Mice and Rice (25)

时间限制

30 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_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:

11 325 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

#include <iostream>#define SIZE 2002using namespace std;int order[SIZE];int order1[SIZE];int rank1[SIZE];int mice[SIZE];int c[20]; //log2 2048 equals to 11////////////////////////////////////////////////////////////////// THE CODEING SPEED IS TOOOOOOO SLOW////////////////////////////////////////int main(){int NP,NG;int i,tmpindex;//freopen("test.txt","r",stdin);scanf("%d%d",&NP,&NG);for(i=0;i<NP;i++)scanf("%d",&mice[i]);for(i = 0;i<NP;i++)        scanf("%d",&order[i]);int level = 1;int size = NP;int *v1 = order1,*v = order;int count,count1=0;while(size != 1){    count1=0;for(i = 0;i+NG-1 < size;i +=NG){tmpindex = i;            for(int j =i+1;j<=i+NG-1;j++){            if(mice[v[j]] >mice[v[tmpindex]]){            rank1[v[tmpindex]] = level;            tmpindex = j;    }else        rank1[v[j]] = level;    c[level]++;    }v1[count1++]= v[tmpindex];}if(i != size){    tmpindex =i;    for(i++;i<size;i++){                if(mice[v[i]] >mice[v[tmpindex]]){                rank1[v[tmpindex]] = level;                tmpindex = i;        }else            rank1[v[i]] = level;        c[level]++;    }    v1[count1++]= v[tmpindex];}level++;size = count1;int *p = v;v = v1;v1 = p;}rank1[v[0]] = level;c[level]++;int pre=0;for(i=level;i>=0;i--){tmpindex = c[i];c[i] = pre+1;pre +=tmpindex;}for(i=0;i<NP;i++){rank1[i] = c[rank1[i]];}for(i =0;i<NP-1;i++){printf("%d ",rank1[i]);}printf("%d\n",rank1[i]);//fclose(stdin);return 0;}