洛谷 P1309 瑞士轮

题目所在网址：https://www.luogu.com.cn/problem/P1309

题目意思并不难理解，就是让你重复做比赛、排序、比赛、排序...这样的工作。但如果真的用快排(sort())去模拟的话，复杂度O(Rnlogn)是过不了的。这里要针对题目特点去变通一下：很明显，假设参赛选手已经排好序，他们打完一架后，胜利的人们拿出来，他们还是有序的，失败的人们拿出来，他们也是有序的，所以在比赛完之后，下一步完全可用归并操作去代替排序操作，将时间复杂度从O(nlogn)降至O(n)；这样就能顺利AC掉本题。

注意：只是归并，不是归并排序！！只需要merge()，不需要merge_sort()画蛇添足

附上AC代码：

#include<iostream>
#include<algorithm>
#include<fstream>
#include<vector>
using namespace std;

const int MAXN = 2e5 + 5;

class athletes{
public:
    int id, score, power;
    athletes(){
    }
    athletes(int id, int score){
        this->id = id;
        this->score = score;
    }
    athletes(const athletes& a){
        this->id = a.id;
        this->score = a.score;
        this->power = a.power;
    }
};

athletes game[MAXN], winner[MAXN], loser[MAXN];

bool cmp(const athletes& a, const athletes& b){
    if(a.score != b.score)return a.score > b.score;
    return a.id < b.id;
}

void merge(athletes game[], athletes winner[], athletes loser[], int n){
    int i = 1, j = 1, l = 1;
    while(i <= n || j <= n){
        if(i > n){
            game[l++] = loser[j++];
            continue;
        }
        if(j > n){
            game[l++] = winner[i++];
            continue;
        }
        if(cmp(winner[i], loser[j])){
            game[l++] = winner[i++];
        }
        else game[l++] = loser[j++];
    }
}

int main(){
//  ifstream cin("data.txt");
//  ios::sync_with_stdio(false);
    int n, r, q, temp;
    cin >> n >> r >> q;
    for(int i = 1; i <= n * 2; i++){
        game[i].id = i;
        cin >> game[i].score;
    }
    for(int i = 1; i <= n * 2; i++){
        cin >> game[i].power;
    }
    sort(game + 1, game + 2 * n + 1, cmp);
    while(r--){
        for(int i = 1; i < 2 * n; i += 2){
            if(game[i].power > game[i + 1].power){
                game[i].score++;
                winner[(i + 1) / 2] = game[i];
                loser[(i + 1) / 2] = game[i + 1];
            }
            else{
                game[i + 1].score++;
                winner[(i + 1) / 2] = game[i + 1];
                loser[(i + 1) / 2] = game[i];
            }
        }
        merge(game, winner, loser, n);
    }
    cout << game[q].id;
    return 0;
}
//2020.3.10
//每次比赛完后，分胜者组和败者组，两组内已经是顺序的，无需排序，O(n)归并一下就好