PAT-A1056 Mice and Rice 题目内容及题解

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

题目大意

题目给定一组游戏结果，参与游戏的顺序是随机决定的，规则为将若干个游戏结果分组进行匹配。最大的数字获胜并进入下一轮，其余人被淘汰，所有的被淘汰者的排名一致，直到确定最后的获胜者。题目给定游戏数据和初始游戏顺序，要求输出其排名。

解题思路

读入题目所给数据并用结构体存储，并记录游戏顺序；
用vector容器模拟队列，并按照题目要求选取最大值，同时记录排名；
边界条件是队列内仅剩一个数据，则排名为1，退出循环；
输出结果并返回零值。

代码

#include<cstdio>
#include<vector>
#define maxn 1010
using namespace std;
 
struct mouse{
    int weight;
    int rank;
}mice[maxn]; 

int main(){
    int NP,NG;
    int Rat,Group;
    int i,j,k,front,current,u,max;
    vector<int> Seq,temp;
    scanf("%d%d",&NP,&NG);
    for(i=0;i<NP;i++){
        scanf("%d",&mice[i].weight);
    }
    for(i=0;i<NP;i++){
        scanf("%d",&j);
        Seq.push_back(j);
    }
    Rat=NP;
    while(1){
        Group=(Rat+NG-1)/NG;//组数
        front=0;
        temp.clear();
        for(i=0;i<Group;i++){
            u=-1,max=-1;
            for(j=0;j<NG;j++){
                current=Seq[front++];
                mice[current].rank=Group+1;
                if(mice[current].weight>max){
                    max=mice[current].weight;
                    u=current;
                }
                if(front==Rat){
                    break;
                }
            }
            temp.push_back(u);
        }
        Rat=Group;
        Seq=temp;
        if(Rat==1){
            mice[Seq[0]].rank=1;
            break;
        }
    }
    for(i=0;i<NP;i++){
        printf("%d",mice[i].rank);
        if(i<NP-1){
            printf(" ");
        }else{
            printf("\n");
        }
    }
    return 0;
}

运行结果

PAT-A1056 Mice and Rice 题目内容及题解