pat1056. Mice and Rice (25)

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 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

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提交代码

 

题目不难。但是做的时候比较长，需要注意。

原因如下：

1.题意理解问题。题目并没有每次等级的计算公式，但网上的说法是rank=np/ng+(np%ng==0?0:1)+1。不知道为什么，我还以为先分出等级，然后由上到下，每一等级由大到小编号。这里时间耗费较多。

2.代码书写。由于等级的计算公式不是理解，所以写代码开始逻辑并不清楚。本来想先每ng处理，然后最后剩下的单独处理，后来一想，其实只要加判断条件 i<nnp 就可以了。

 

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
#include<string>
#include<map>
#include<set>
using namespace std;
int ra[1005],weight[1005],order[1005];
int main(){
    //freopen("D:\\INPUT.txt","r",stdin);
    int np,ng;
    scanf("%d %d",&np,&ng);
    int i,j;
    for(i=0;i<np;i++){
        scanf("%d",&weight[i]);
    }
    for(i=0;i<np;i++){
        scanf("%d",&order[i]);
    }


    int nnp=np;
    int r=nnp/ng+(nnp%ng==0?0:1)+1;//每局的等级;
    int s,maxnum;
    queue<int> q;//存放编号
    for(i=0;i<nnp;){
        ra[order[i]]=r;
        maxnum=order[i];//每一轮都有最大值
        s=i++;
        for(;i<nnp&&i-s<ng;i++){
            ra[order[i]]=r;
            if(weight[maxnum]<weight[order[i]]){
                maxnum=order[i];
            }
        }
        q.push(maxnum);//每组的最大值进入到下一次比赛中
        //cout<<maxnum<<endl;
    }
    while(q.size()>1){
        nnp=q.size();
        r=nnp/ng+(nnp%ng==0?0:1)+1;
        for(i=0;i<nnp;){
            maxnum=q.front();
            q.pop();
            ra[maxnum]=r;
            s=i++;
            for(;i<nnp&&i-s<ng;i++){
                ra[q.front()]=r;
                if(weight[maxnum]<weight[q.front()]){
                    maxnum=q.front();
                }
                q.pop();//不要忘
            }
            q.push(maxnum);
        }
    }
    ra[q.front()]=1;
    printf("%d",ra[0]);
    for(i=1;i<np;i++){
        printf(" %d",ra[i]);
    }
    return 0;
}