（难+ 要再看）A1056 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 (≤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

题意：

给定 NP 只老鼠的质量，并给出他们的初始序列，按照这个初始序列把这些老鼠按每 NG 只分为一组，最后不够 NG 只的也单独分为一组。对每组老鼠，选出他们中质量最大的一只晋级，这样晋级的；老鼠就等于该轮分组的组数，对这些晋级的老鼠再按照上面的步骤每 NG 只分为一组进行比较，选出质量最大的一批继续晋级，这样直到最后只剩下 1 只老鼠，排名为 1。把这些老鼠的排名按照原输入的顺序输出。

输入：第一行给出 NP 和 NG (≤1000)，第二行按序给出每只老鼠的质量，第三行给出初始序列（即 6 号老鼠排在第一位，0 号老鼠排在第二位：也就是 19 在第一个，25 在第二个）

样例分析：

有输入可以得到，初始时老鼠的顺序为：

第一次 19 25 57 22 10 3 56 18 37 0 46

第二次 57 56 46

第三次 57 46

第四次 57

知识点：

queue

添加 #include <queue>; using namespace std;
定义：queue<typename> name;
队列只能通过 front() 和 back() 来访问首尾元素
其他用法基本与 stack 相同

运行超时

#include <cstdio>
#include <queue>
using namespace std;
const int maxn = 1010;
struct mouse{
    int weight;
    int R;
}mouse[maxn];
int main(){
    int nq, ng, order;
    scanf("%d%d", &nq, &ng);
    for(int i = 0; i < nq; i++){
        scanf("%d", &mouse[i].weight);
    }
    queue<int> q;
    for(int i = 0; i < nq; i++){
        scanf("%d", &order);
        q.push(order);      //按顺序把老鼠们的标号入队
    }
    int temp = nq, group;   //temp为当前轮的比赛总老鼠数，group为组数
    while(q.size() != 1){
        //计算group，即当前轮分为几组进行比较
        if(temp % nq == 0)
            group = temp / nq;
        else
            group = temp / nq + 1;
        //枚举每一组，选出该组老鼠中质量最大的
        for(int i = 0; i < group; i++){
            int k = q.front();      //k 存放该组质量最大的老鼠编号
            for(int j = 0; j < ng; j++){
                if(i * ng + j >= temp)
                    break;
                int front = q.front();
                if(mouse[front].weight > mouse[k].weight){
                    k = front;
                }
                mouse[front].R = group + 1;
                q.pop();
            }
            q.push(k);
        }
        temp = group;
    }
    mouse[q.front()].R = 1;
    for(int i = 0; i < nq; i++){
        printf("%d", mouse[i].R);
        if(i < nq - 1)
            printf(" ");
    }
    return 0;
}