1056 Mice and Rice

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:

?

1 2 3

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:

?

1	5 5 5 2 5 5 5 3 1 3 5

 

思路 

首先可以简单证明，当前老鼠分为group组时，被淘汰的老鼠的排名是group+1。

这样不断淘汰直到剩下一只老鼠，比赛结束。题目增加了难度，老鼠不是依次比赛，而是按照题目给的序列，进行比赛。

我们使用一个队列来依次存放老鼠对应的下标。

 

核心代码：

 

while((cnt = (int)q.size()) > 1){
        group = cnt / g + (cnt % g == 0 ? 0 : 1);
        for(int i = 0; i < group; i++){
            mx = -1;
            mxi = -1;
            for(int j = 0; j < g && i * g + j < cnt; j++){
                r[q.front()] = group + 1;
                if(w[q.front()] > mx){
                    mx = w[q.front()];
                    mxi = q.front();
                }
                q.pop();
            }
            q.push(mxi);
        }
    }

 

 

 

 

代码

#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std;
int w[1005],r[1005];
int main(){
    int p,g,o;
    queue<int>q;
    cin>>p>>g;
    for(int i=0; i<p; i++){
        cin>>w[i];
    }
    for(int i=0; i<p; i++){
        cin>>o;
        q.push(o);
    }
    int group, cnt, mx, mxi;
    while((cnt = (int)q.size()) > 1){
        group = cnt / g + (cnt % g == 0 ? 0 : 1);
        for(int i = 0; i < group; i++){
            mx = -1;
            mxi = -1;
            for(int j = 0; j < g && i * g + j < cnt; j++){
                r[q.front()] = group + 1;
                if(w[q.front()] > mx){
                    mx = w[q.front()];
                    mxi = q.front();
                }
                q.pop();
            }
            q.push(mxi);
        }
    }
    r[q.front()] = 1;
    printf("%d", r[0]);
    for(int i = 1; i < p; i++){
        printf(" %d", r[i]);
    }
    return 0;
}