PAT 1056 Mice and Rice[难][不理解]

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 (≤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：

//看了一遍题意，愣是没看懂。看了三遍还是没看懂，放弃了。看了题解上说的题意，还不不太明白，算了吧，看代码吧。

代码来自： https://www.liuchuo.net/archives/2936

#include <iostream>

#include <queue>

#include <vector>

#include <algorithm>

using namespace std;

struct node {

    int weight, index, rank, index0;

};

bool cmp1(node a, node b) {

    return a.index0 < b.index0;

}

int main() {

    int n, g, num;

    scanf("%d%d", &n, &g);

    vector<int> v(n);

    vector<node> w(n);

    for(int i = ; i < n; i++)

        scanf("%d", &v[i]);

    for(int i = ; i < n; i++) {

        scanf("%d", &num);

        w[i].weight = v[num];//这个存的是体重

        w[i].index = i;//这个存的是第几个，这个是对应老鼠编号的。

        w[i].index0 = num;//这个存的是初始顺序。

    }

    queue<node> q;

    for(int i = ; i < n; i++)

        q.push(w[i]);

    while(!q.empty()) {//目的是找出最胖的。

        int size = q.size();

        if(size == ) {

            node temp = q.front();

            w[temp.index].rank = ;

            break;

        }

        int group = size / g;

        if(size % g != )//这样来安排最后鼠数几个不够g的。

            group += ;

        node maxnode;

        int maxn = -, cnt = ;

        for(int i = ; i < size; i++) {

            node temp = q.front();

            w[temp.index].rank = group + ;

            q.pop();

            cnt++;

            if(temp.weight > maxn) {

                maxn = temp.weight;

                maxnode = temp;

            }

            if(cnt == g || i == size - ) {

                cnt = ;

                maxn = -;

                q.push(maxnode);

            }

        }

    }

    sort(w.begin(), w.end(), cmp1);

    for(int i = ; i < n; i++) {

        if(i != ) printf(" ");

        printf("%d", w[i].rank);

    }

    return ;

}

//还是不太懂什么意思，以后再说。

巴特西