PAT-1056 Mice and Rice （分组决胜问题）

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 N_P programmers. Then every N_Gprogrammers 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

题目大意：这是一个比赛中不断分组最终决出胜者的题目。首先给出选手数量np和每组人数ng，然后按序号顺序给出每个选手的分数，接着是以选手序号组成的比赛顺序，按照这个顺序把选手分成不同组，每组有一个优胜者加入下一轮并重复上述分组，而失败者拥有相同的排名并被淘汰。要求输出最终所有选手的排名。

主要思想：思路很简单，利用循环来模拟每轮的比赛，首先需要求出当前轮的组数，然后对于每组成员找出得分最高的选手并将其加入下一轮，将失败者的排名设置为当前组数+1。其中重点是如何更新下一轮的比赛顺序，这里用的是vector容器来储存比赛顺序，将每一组胜者直接覆盖在数组的前端，而下一轮比赛的选手数就是此轮的组数。

#include <iostream>

#include <vector>

using namespace std;

int weight[1000];			//选手得分

int rank_id[1000];			//选手排名

vector<int> order;			//比赛顺序

int get_max(int lo, int hi, int p);

int min(int a, int b);

int main(void) {

    int np, ng, i;

    cin >> np >> ng;

    for (i = 0; i < np; i++)  {

        cin >> weight[i];

    }

    for (i = 0; i < np; i++) {

        int t;

        cin >> t;

        order.push_back(t);

    }

    int n = np;

    while (true) {

        int groups = n / ng;						//n是当前轮比赛的人数

        int r = n % ng;

        if (r > 0)      groups++;					//求出分组数

        int lo = 0, count = 0;

        int win_id;

        for (i = 0; i < groups; i++) {

            int hi = min(lo+ng-1, n-1);

            win_id = get_max(lo, hi, groups);		//该组的优胜者序号

            order[count++] = win_id;				//不断更新下一轮的

            lo = hi+1;

        }

        n = groups;

        if (n == 1) {

            rank_id[win_id] = 1;

            break;

        }

    }

    cout << rank_id[0];

    for (i = 1; i < np; i++)

        cout << " " << rank_id[i];

    cout << endl;

    return 0;

}

//获得当前组优胜选手的序号，并且把失败者排名设置为group+1

int get_max(int lo, int hi, int p) {

    int max_id, i;

    int max = 0;

    for (i = lo; i <= hi; i++) {

        if (max < weight[order[i]]){

            max = weight[order[i]];

            max_id = order[i];

        }

    }

    for (i = lo; i <= hi; i++) {

        if (order[i] != max_id)

            rank_id[order[i]] = p + 1;

    }

    return max_id;

}

int min(int a, int b) {

    return a < b ? a : b;

}

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PAT-1056 Mice and Rice （分组决胜问题）

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