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  • 1056 Mice and Rice (25 分)

    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​​ (≤), 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​​ (,) 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 (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
     
     本题应该是之前的保研机试题,使用队列,具体逻辑很复杂,后期再修改
    #include<bits/stdc++.h>
    using namespace std;
    const int maxn=1010;
    #define inf 0x3fffffff
    struct Node{
        int w;
        int rank;
    }mouse[maxn];
    int main(){
        int np,ng,order;
        scanf("%d %d",&np,&ng);
        for(int i=0;i<np;i++){
            scanf("%d",&mouse[i].w);
        }
        queue<int> q;
        for(int i=0;i<np;i++){
            scanf("%d",&order);
            q.push(order);
        }
        int temp=np,group;
        while(q.size()!=1){
            if(temp%np==0){
                group=temp/np;
            }
            else{
                group=temp/np+1;
            }
            for(int i=0;i<group;i++){
                int k=q.front();
                for(int j=0;j<ng;j++){
                    if(i*ng+j>=temp){
                        break;
                    }
                    int front=q.front();
                    if(mouse[front].w>mouse[k].w){
                        k=front;
                    }
                    mouse[front].rank=group+1;
                    q.pop();
                }
                q.push(k);
            }
            temp=group;
        }
        mouse[q.front()].rank=1;
        for(int i=0;i<np;i++){
            printf("%d",mouse[i].rank);
            if(i<np-1){
                printf(" ");
            }
        }
        printf("
    ");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dreamzj/p/15032759.html
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