PAT甲级2018春真题

算法字符串哈希排序图论

字数统计: 874阅读时长: 5 min

 2023/09/10 

PAT考试真题解析

[1140] Look-and-say Sequence (20)

字符串

题目

代码

int main() {
  string s;
  int n, j;
  cin >> s >> n;
  for (int cnt = 1; cnt < n; cnt++) {
    string t;
    for (int i = 0; i < s.length(); i = j) {
      for (j = i; j < s.length() && s[j] == s[i]; j++);
        t += s[i] + to_string(j - i);
      }
    s = t;
  }
  cout << s;
  return 0;
}

[1141] PAT Ranking of Institutions (25)

哈希、排序

题目

代码

unordered_map<string, double> weighted_score;
unordered_map<string, int> num;

int main() {
  double score;
  int rank = 0, n;
  cin >> n;
  string uid, school;
  for (int i = 0; i < n; i++) {
    cin >> uid >> score >> school;
    for (int j = 0; j < school.length(); j++)
      school[j] = tolower(school[j]);
    num[school]++;
    if (uid[0] == 'A') {
      weighted_score[school] += score;
    } else if (uid[0] == 'T') {
      weighted_score[school] += score * 1.5;
    } else {
      weighted_score[school] += score * 1.0 / 1.5;
    }
  }

  vector<pair<string, int>> ans;
  for (auto it = weighted_score.begin(); it != weighted_score.end(); it++)
    ans.push_back(make_pair(it -> first, (int)(it -> second)));
  sort(ans.begin(), ans.end(), [&](auto a, auto b) {
    return a.second == b.second? (num[a.first] == num[b.first]? a.first < b.first: num[a.first] < num[b.first]): a.second > b.second;
  });

  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    if (i && ans[i].second == ans[i-1].second) {}
    else  rank = i + 1;
    printf("%d %s %d %d\n", rank, ans[i].first.c_str(), (int)ans[i].second, num[ans[i].first]);
  }
  return 0;
}

[1142] Maximal Clique (25)

图论

题目

代码

vector<vector<int>> G;
vector<int> vis;

int judge(vector<int> &clique, int n) {
  for (int i = 0; i < clique.size() - 1; i++) {
    for (int j = i + 1; j < clique.size(); j++) {
      if (G[clique[i]][clique[j]] != 1)
        return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    int cnt = 0;
    for (auto q: clique)
      if (G[i][q] == 1) cnt++;
    if (cnt == clique.size()) return 1;
  }
  return 2;
}

int main() {
  int n, m, a, b, k;
  cin >> n >> m;
  G.resize(n + 1, vector<int>(n + 1));
  vis.resize(n + 1);
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    G[a][b] = 1;
    G[b][a] = 1;
  }
  cin >> k;
  for (int i = 0; i < k; i++) {
    cin >> a;
    vector<int> clique;
    fill(vis.begin(), vis.end(), false);
    for (int j = 0; j < a; j++) {
      cin >> b;
      clique.push_back(b);
      vis[b] = true;
    }
    int ans = judge(clique, n);
    if (ans == 2) printf("Yes\n");
    else if (ans == 1)  printf("Not Maximal\n");
    else  printf("Not a Clique\n");
  }
  return 0;
}

[1143] Lowest Common Ancestor (30)

题目

注意点与解析

map<int, bool> mp ⽤来标记树中所有出现过的结点，遍历⼀遍pre数组，将当前结点标记为a，如果u和v
分别在a的左、右，或者u、v其中⼀个就是当前a，即 (a >= u && a <= v) || (a >= v && a <= u)，说明找到了这个共同最低祖先a，退出当前循环，最后根据要求输出结果即可

代码

map<int, bool> mp;
int main() {
  int m, n, u, v, a;
  scanf("%d %d", &m, &n);
  vector<int> pre(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &pre[i]);
    mp[pre[i]] = true;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &u, &v);
    for(int j = 0; j < n; j++) {
      a = pre[j];
      if ((a >= u && a <= v) || (a >= v && a <= u)) break;
    }
    if (mp[u] == false && mp[v] == false)
      printf("ERROR: %d and %d are not found.\n", u, v);
    else if (mp[u] == false || mp[v] == false)
      printf("ERROR: %d is not found.\n", mp[u] == false ? u : v);
    else if (a == u || a == v)
      printf("%d is an ancestor of %d.\n", a, a == u ? v : u);
    else
      printf("LCA of %d and %d is %d.\n", u, v, a);
  }
  return 0;
}

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CATALOG

1. [1140] Look-and-say Sequence (20)
1. 1.1. 题目
2. 1.2. 代码
2. [1141] PAT Ranking of Institutions (25)
1. 2.1. 题目
2. 2.2. 代码
3. [1142] Maximal Clique (25)
1. 3.1. 题目
2. 3.2. 代码
4. [1143] Lowest Common Ancestor (30)