A1056. 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 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

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
typedef struct node{
    int weight;
    int rank;
    int rank_finall;
}info;
info mice[100001];
queue<int> qu;
int rk[100001];
bool cmp(int a, int b){
    return mice[a].rank > mice[b].rank;
}
int main(){
    int NP, NG, comp, temp;
    scanf("%d%d", &NP, &NG);
    for(int i = 0; i < NP; i++){
        scanf("%d", &mice[i].weight);
        mice[i].rank = 0;
        rk[i] = i;
    }
    for(int i = 0; i < NP; i++){
        scanf("%d", &temp);
        qu.push(temp);
    }
    comp = NP;  //该轮比赛的人数
    int rank = 1;
    while(comp > 1){
        for(int j = 0; j < comp; j += NG){
            int maxW = -1;
            int maxIndex, tempIndex;
            for(int k = 0; k < NG && j + k < comp; k++){  //防止最后一轮人数不足
                tempIndex = qu.front();
                mice[tempIndex].rank = rank;
                qu.pop();
                if(mice[tempIndex].weight > maxW){
                    maxW = mice[tempIndex].weight;
                    maxIndex = tempIndex;
                }
            }
            qu.push(maxIndex);
        }
        rank++;
        comp = comp % NG == 0 ? comp / NG : comp / NG + 1;
    }
    int maxPt = qu.front();
    mice[maxPt].rank = rank;
    sort(rk, rk + NP, cmp);
    mice[rk[0]].rank_finall = 1;
    for(int i = 1; i < NP; i++){
        if(mice[rk[i]].rank == mice[rk[i - 1]].rank)
            mice[rk[i]].rank_finall = mice[rk[i - 1]].rank_finall;
        else mice[rk[i]].rank_finall = i + 1;
    }
    printf("%d", mice[0].rank_finall);
    for(int i = 1; i < NP; i++){
        printf(" %d", mice[i].rank_finall);
    }
    cin >> NP;
    return 0;
}

总结：

1、题意：给出NP只老鼠比体重大小，比法是：每NG个分为一组选出最大的进入下一轮，下一轮依旧每NG各分一组......直到选出最大的。题目要求按照输入的老鼠的顺序输出他们的排名（并列的老鼠名次相同，但占用排位）。另外注意给老鼠分组时，最后一组不满NG个的情况但仍然算一组。

2、queue、stack、vector等存储的是数据的拷贝而不是引用，所以在对struct数据使用这些容器时，最好的办法是将所有的struct存储在data数组中，而将它们的数组下标作为vector等的元素，相当于存储了引用。

3、最好使用STL的queue而非自己写一个队列。