2021 ICPC网络赛第一场

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2021 ICPC网络赛第一场

A

• 题意

  有$k$个点（ $j=0,1...k$），$n$个任务（$i=0,1...n$），每个任务有到达时间和持续时间，第$i$个任务从$i\%k$的点开始往后找有没有空闲的点，如果没有找到则任务作废，问$k$个点谁接的任务最多

• 思路

  如果暴力查找，从第二个点开始找，找完一遍发现是第一个点，碰到多个这种数据这样肯定会$T$，所以考虑在查找方面减少复杂度，这里采取线段树+二分+set

  • 线段树维护$k$个点结束的区间最小值，单点修改，区间查询
  • 二分查找最小值
  • $set$维护$k$个点结束的最小值，判断是否任务需要遗弃

• 代码

  /*
   * @Author: NEFU AB-IN
   * @Date: 2021-09-19 13:03:13
   * @FilePath: \Contest\a.cpp
   * @LastEditTime: 2021-09-19 19:27:05
   */
  #include 
  using namespace std;
  
  const int N = 1e5 + 10;
  int a[N << 2], tr[N << 3], cnt[N];
  set<int> s;
  map<int, int> m;
  
  void bulid(int i, int l, int r)
  {
      if (l == r)
      {
          tr[i] = 0x3f3f3f3f;
          return;
      }
      int mid = l + r >> 1;
      bulid(i << 1, l, mid);
      bulid(i << 1 | 1, mid + 1, r);
      tr[i] = min(tr[i << 1], tr[i << 1 | 1]);
  }
  
  void update(int i, int l, int r, int x, int y)
  {
      if (l > x || x > r)
          return;
      if (l == r && l == x)
      {
          tr[i] = y;
          return;
      }
      int mid = l + r >> 1;
      update(i << 1, l, mid, x, y);
      update(i << 1 | 1, mid + 1, r, x, y);
      tr[i] = min(tr[i << 1], tr[i << 1 | 1]);
  }
  
  int query(int i, int l, int r, int x, int y)
  {
      if (y < l || x > r)
          return 0x3f3f3f3f;
      if (l >= x && r <= y)
          return tr[i];
      int mid = l + r >> 1;
      return min(query(i << 1, l, mid, x, y), query(i << 1 | 1, mid + 1, r, x, y));
  }
  
  int f(int l, int r, int x, int k)
  {
      if (l == r)
          return l;
      int mid = (l + r) / 2;
      if (query(1, 1, 2 * k, l, mid) < x)
          f(l, mid, x, k);
      else
          f(mid + 1, r, x, k);
  }
  
  int main()
  {
      int k, n;
      cin >> k >> n;
      bulid(1, 1, 2 * k);
      s.insert(0);
      m[0] = k;
      for (int i = 0; i < n; ++i)
      {
          int l, time;
          cin >> l >> time;
          if (*s.begin() >= l)
              continue;
          if (a[i % k] < l)
          {
              cnt[i % k]++;
              m[a[i % k]]--;
              if (m[a[i % k]] == 0)
                  s.erase(a[i % k]);
              a[i % k] = l + time - 1;
              if (s.find(a[i % k]) == s.end())
                  s.insert(a[i % k]);
              m[a[i % k]]++;
              update(1, 1, 2 * k, (i % k) + 1, l + time - 1);
              update(1, 1, 2 * k, (i % k) + k + 1, l + time - 1);
          }
          else
          {
              int id = f(i % k + 1, i % k + k, l, k);
              id--;
              id %= k;
              cnt[id]++;
              m[a[id]]--;
              if (m[a[id]] == 0)
                  s.erase(a[id]);
              a[id] = l + time - 1;
              if (s.find(a[id]) == s.end())
                  s.insert(a[id]);
              m[a[id]]++;
              update(1, 1, 2 * k, id + 1, l + time - 1);
              update(1, 1, 2 * k, id + k + 1, l + time - 1);
          }
      }
      int minx = 0, ans = 0;
      for (int i = 0; i < k; i++)
          minx = max(minx, cnt[i]);
      for (int i = 0; i < k; i++)
      {
  
          if (cnt[i] == minx)
          {
              if (ans)
                  printf(" ");
              printf("%d", i);
              ans++;
          }
      }
      return 0;
  }
  

F

• 题意

  几何

• 代码

  /*
   * @Author: NEFU AB-IN
   * @Date: 2021-09-19 12:04:27
   * @FilePath: \Contest\f.cpp
   * @LastEditTime: 2021-09-19 12:28:19
   */
  #include 
  using namespace std;
  
  const int N = 1e5 + 10;
  int a[N];
  
  int main()
  {
      int t;
      cin >> t;
      for (int i = 1; i <= t; ++i)
      {
          double a, b, r;
          cin >> a >> b >> r;
          if (r >= b)
          {
              printf("Case #%d: %.2f\n", i, 2 * a - r);
          }
          else
          {
              printf("Case #%d: %.2f\n", i, 2 * sqrt(a * a + (b - r) * (b - r)) - r);
          }
      }
      return 0;
  }
  

H

• 题意

  队友出的，贴一下

• 代码

  /*
   * @Author: NEFU AB-IN
   * @Date: 2021-09-19 14:44:27
   * @FilePath: \Contest\h.cpp
   * @LastEditTime: 2021-09-19 15:01:14
   */
  #include 
  using namespace std;
  const int N = 1e5 + 5;
  int n, m;
  set<int> s1[N], s2[N];
  map<int, int> map1;
  void insert111(int a, int b)
  {
      if (s1[map1[a]].find(b) == s1[map1[a]].end())
          s1[map1[a]].insert(b);
  }
  int main()
  {
      ios::sync_with_stdio(0);
      cin.tie(0);
      cout.tie(0);
      cin >> n >> m;
      int id, biao, a, b, c;
      double x, y, z;
      for (int i = 1; i <= n; i++)
      {
          cin >> id;
          map1[id] = i;
          cin >> x >> y >> z;
          // printf("%lf %lf %lf\n", x, y, z);
      }
      for (int i = 1; i <= m; i++)
      {
          cin >> id >> biao;
          switch (biao)
          {
          case 102:
              cin >> a >> b;
              insert111(a, b);
              s2[map1[a]].insert(id);
              insert111(b, a);
              s2[map1[b]].insert(id);
              break;
          case 203:
              cin >> a >> b >> c;
              insert111(a, b);
              insert111(a, c);
              insert111(b, c);
              insert111(b, a);
              insert111(c, a);
              insert111(c, b);
              s2[map1[a]].insert(id);
              s2[map1[b]].insert(id);
              s2[map1[c]].insert(id);
          }
      }
      int t;
      cin >> t;
      while (t--)
      {
          cin >> id;
          if (map1[id] == 0)
          {
              printf("%d\n[]\n[]\n", id);
          }
          else
          {
              printf("%d\n[", id);
              for (auto i : s1[map1[id]])
              {
                  if (i != *s1[map1[id]].begin())
                      printf(",");
                  printf("%d", i);
              }
              printf("]\n[");
              for (auto i : s2[map1[id]])
              {
                  if (i != *s2[map1[id]].begin())
                      printf(",");
                  printf("%d", i);
              }
              printf("]\n");
          }
      }
      return 0;
  }
  

I

• 题意

  签到题，数据范围也不给，考验选手读入能力，点赞

• 代码

  '''
  Author: NEFU AB-IN
  Date: 2021-09-19 12:10:49
  FilePath: \Contest\i.py
  LastEditTime: 2021-09-19 12:12:43
  '''
  lst = list(map(int, input().split()))
  x = int(input())
  r = int(input())
  lst.sort()
  flag = 0
  for i in range(len(lst) - 1, -1, -1):
      if abs(lst[i] - x) <= r:
          print(lst[i], end=" ")
          flag = 1
  if flag == 0:
      print()
  

K

• 题意

  模拟即可

• 代码

   /*
  * @Author: NEFU AB-IN
  * @Date: 2021-09-19 15:03:00
  * @FilePath: \Contest\k.cpp
  * @LastEditTime: 2021-09-21 20:26:23
  */
  #include 
  using namespace std;
  int n, m;
  const int N = 1e5 + 50;
  vector<int> v[N];
  
  int main()
  {
     int t;
     scanf("%d", &t);
     for (int kk = 1; kk <= t; kk++)
     {
         scanf("%d%d", &n, &m);
         int xx, u;
         for (int i = 1; i <= n; i++)
         {
             v[i].clear();
             cin >> xx;
             for (int j = 1; j <= xx; j++)
             {
                 scanf("%d", &u);
                 v[i].push_back(u);
             }
         }
         printf("Case #%d: \n", kk);
         for (int j = 1; j <= m; j++)
         {
             bool flag = 0;
             int k;
             scanf("%d %d", &u, &k);
             int vv = u, id;
             for (int i = 1; i <= k; i++)
             {
                 scanf("%d", &id);
                 if (id > v[vv].size() || id == 0)
                     flag = 1;
                 if (flag == 1)
                     continue;
  
                 vv = v[vv][id - 1];
             }
             if (flag)
                 printf("Packet Loss");
             else
             {
                 printf("%d", vv);
             }
             if (j != m)
                 printf("\n");
         }
         if (kk != t)
             printf("\n");
     }
     return 0;
  }
  

明天补题。。