摘要
Title: L2-009 抢红包 (25 分)
Tag: 模拟
Memory Limit: 64 MB
Time Limit: 1000 ms

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• 题意
• 思路
• 代码

L2-009 抢红包 (25 分)

• 题意

  没有人没抢过红包吧…… 这里给出N个人之间互相发红包、抢红包的记录，请你统计一下他们抢红包的收获。

• 思路

  模拟即可

• 代码

  /*
  * @Author: NEFU AB-IN
  * @Date: 2022-04-20 20:47:30
  * @FilePath: \ACM\GPLT\L2-009.CPP
  * @LastEditTime: 2022-04-20 20:56:41
  */
  #include <bits/stdc++.h>
  using namespace std;
  #define int long long
  #define MP make_pair
  #define SZ(X) ((int)(X).size())
  #define IOS                                                                                                            \
      ios::sync_with_stdio(false);                                                                                       \
      cin.tie(0);                                                                                                        \
      cout.tie(0);
  #define DEBUG(X) cout << #X << ": " << X << endl;
  typedef pair<int, int> PII;
  const int INF = 0x3f3f3f3f;
  
  const int N = 1e4 + 10;
  int n;
  int cnt[N];
  double sum[N];
  
  struct sa
  {
      int id, cnt;
      double sum;
  };
  
  signed main()
  {
      IOS;
      cin >> n;
      for (int i = 1; i <= n; ++i)
      {
          int k;
          double sum1 = 0.0;
          cin >> k;
          for (int j = 1; j <= k; ++j)
          {
              int v;
              double p;
              cin >> v >> p;
              cnt[v] += 1;
              sum[v] += p;
              sum1 += p;
          }
          sum[i] -= sum1;
      }
      vector<sa> v;
      for (int i = 1; i <= n; ++i)
      {
          v.push_back({i, cnt[i], sum[i]});
      }
  
      sort(v.begin(), v.end(), [&](sa a, sa b) {
          if (a.sum != b.sum)
              return a.sum > b.sum;
          else if (a.cnt != b.cnt)
              return a.cnt > b.cnt;
          return a.id < b.id;
      });
  
      for (auto [id, cnt_, sum_] : v)
      {
          printf("%lld %.2lf\n", id, sum_ / 100.0);
      }
      return 0;
  }