PAT 1025 PAT Ranking

最新推荐文章于 2024-01-13 16:20:35 发布

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最新推荐文章于 2024-01-13 16:20:35 发布

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分类专栏： PAT

本文链接：https://blog.youkuaiyun.com/qq_25185969/article/details/113184301

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本文介绍了一种用于处理PAT竞赛中选手排名的算法实现。该算法通过局部排序和全局排序两个阶段来确定每位参赛者的最终名次，并考虑了分数相同情况下的处理方式。通过优先队列进行高效管理，确保了排名的准确性和快速生成。

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#include <cmath>
#include <vector>
#include <climits>
#include <iostream>
#include <string>
#include <map>
#include <queue>
#include <algorithm>

using namespace std;

class Testee {
public:
  string id;
  int score;
  int location;
  int localRank;
  int totalRank;

  friend bool operator < (Testee t1, Testee t2) {
    if (t1.score != t2.score) return t1.score < t2.score;
    else return t1.id > t2.id;
  }
};

bool compare1025(Testee t1, Testee t2) {
  return t1 < t2;
}

int main() {

  int N, K, totalSize = 0;
  cin >> N;

  vector<vector<Testee>> data(N);
  // 1. 读入数据
  for (int location = 1; location <= N; location++) {
    cin >> K;
    for (int i = 1; i <= K; i++) {
      Testee testee;
      cin >> testee.id >> testee.score;
      testee.location = location;
      data[location - 1].push_back(testee);
    }
    totalSize += K;
  }

  // 2. 局部排序
  priority_queue<Testee> heap;
  for (int location = 0; location < N; location++) {
    vector<Testee>& list = data[location];
    sort(list.begin(), list.end(), compare1025);
    int lastScore = INT_MAX;
    int size = list.size();
    for (int i = size - 1; i >= 0; i--) {
      list[i].localRank = list[i].score == lastScore ? list[i + 1].localRank : (size - i);
      lastScore = list[i].score;
      if (i == size - 1) {
        heap.push(list[i]);
        list.pop_back();
      }
    }
  }

  // 3. 全局排序，赋予全局序号
  Testee prev;
  cout << totalSize << endl;
  int rank = 1;
  while (!heap.empty()) {
    Testee testee = heap.top();
    testee.totalRank = prev.score == testee.score ? prev.totalRank : rank;
    cout << testee.id << " " << testee.totalRank << " " << testee.location << " " << testee.localRank << endl;

    prev = testee;
    rank += 1;

    heap.pop();
    vector<Testee> &list = data[testee.location - 1];
    if (list.size() > 0) {
      heap.push(list[list.size() - 1]);
      list.pop_back();
    }
  }

  return 0;
}

全局排序那里直接将data[i]合并后排序亦可；可见PAT对时间复杂度要求并不高，考试时无需想太多，sort当用则用， $k \times n ^ 2$ 的复杂度妥妥可以接受