PAT甲级1056 Mice and Rice （25 分）

/*
题意难理解
用样例来解释
11 3
25 18 0 46 37 3 19 22 57 56 10
6 0 8 7 10 5 9 1 4 2 3
Np = 11，表示老鼠的数量
Ng = 3，表示比较的时候每3个老鼠一组
weight[]:25 18 0 46 37 3 19 22 57 56 10，第i个老鼠的重量是weight[i]
6 0 8 7 10 5 9 1 4 2 3，比较的顺序是这样的，weight[6],weight[0],weight[8]是一组进行比较，7,10,5一组，9,1,4一组，2,3一组

还有一点需注意：
Rank[i] = i输掉的那一轮里胜出者的数量+1
*/
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1002;
int Np, Ng;
int weight[maxn];
int Rank[maxn];
queue<int> win, lose;///win记录胜者
int main(){
    scanf("%d%d", &Np, &Ng);
    for(int i = 0; i < Np; ++i)
        scanf("%d", &weight[i]);
    for(int i = 0; i < Np; ++i){
        int d;
        scanf("%d", &d);
        win.push(d);
    }
    while(!win.empty()){
        int winAmount = win.size();///当前胜者有多少，须记录一下，不然下面pop之后win.size()会改变
        if(winAmount == 1){///重点！如果当前只剩下一个胜者，记录一下break即可，否则会死循环
            Rank[win.front()] = 1;
            break;
        }
        int round = (winAmount + Ng - 1) / Ng;///winAmount个胜者再次比较需要多少轮

        for(int i = 0; i < round; ++i){
            int maxWeightNum = win.front();
            for(int j = 0; j < Ng && (i*Ng+j) < winAmount; ++j){///(i*Ng+j) < winAmount来协助判断，防止访问到队列里下一轮的，这也是前面设置winAmount的原因
                if(weight[win.front()] > weight[maxWeightNum]){
                    maxWeightNum = win.front();///筛选这一轮这一组的最大重量的老鼠
                }
                lose.push(win.front());///无论是否是胜者，都先入队列lose，出队列win
                win.pop();
            }
            win.push(maxWeightNum);///胜者再次入win队列
            while(!lose.empty()){
                if(lose.front() != maxWeightNum){///非胜者填写Rank
                    Rank[lose.front()] = round+1;
                }
                lose.pop();
            }
        }
    }
    printf("%d", Rank[0]);
    for(int i = 1; i < Np; ++i){
        printf(" %d", Rank[i]);
    }
    printf("\n");
    return 0;
}