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 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

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。