昨晚的cf打得非常抽象，成功从蓝掉到青，不过在心理预期范围内，可以接受，之后每一场都会跟着打大号，能力的提升比表面上的分数更加重要

CF 1922A Tricky Template

题目链接

太久没训了读题能力飞速下滑，昨晚愣是看了好久才看明白题目是啥意思，加上大脑停转最后写了个抽象的模拟

#include <bits/stdc++.h>using namespace std;#define int long longusing i64 = long long;typedef pair<int, int> PII;void solve()
{int n; cin >> n;string a, b, c;cin >> a >> b >> c;if (a == c || b == c){cout << "NO\n";return;}string ans = "";for (int i = 0; i < n; i ++ ){if (a[i] == b[i]) ans += a[i];else{if (a[i] != c[i] && b[i] != c[i]) ans += c[i] - 'a' + 'A';else ans += "!";}}bool flag = true;for (int i = 0; i < n; i ++ ){if (ans[i] - 'a' < 0){ans[i] = 'a' + ans[i] - 'A';if (ans[i] == c[i]){cout << "YES\n";return;}}else{if (ans[i] != c[i]){cout << "YES\n";return;}}}cout << "NO\n";
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1922B Forming Triangles

题目链接

又是抽象的错误点，没看到二的次方qaq
当较小的两条边是2^x和2^y时，x<y，那么最大的边一定是2^y

#include <bits/stdc++.h>using namespace std;#define int long longusing i64 = long long;typedef pair<int, int> PII;void solve()
{int n;cin >> n;vector<int> a(n);map<int, int> mp;for (int i = 0; i < n; i ++ ){cin >> a[i];mp[a[i]] ++ ;}sort(a.begin(), a.end());int ans = 0;mp[a[0]] -- ;for (int i = 1; i < n; i ++ ){mp[a[i]] -- ;ans += mp[a[i]] * i;}cout << ans << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1922C Closest Cities

题目链接

前两题wa太多次导致写这一题开始急了上手就是个Dijkstra，于是成功t了两发，于是反手就是两个Dijkstra，赛后才发现只有相邻的点有边，非常喜欢小题大做的一个人…

#include <bits/stdc++.h>using namespace std;#define int long longusing i64 = long long;typedef pair<int, int> PII;void solve()
{int n;cin >> n;vector<int> a(n);for (int i = 0; i < n; i ++ ) cin >> a[i];vector<vector<PII>> g(n);g[0].push_back(make_pair(1, 1));g[n - 1].push_back(make_pair(n - 2, 1));for (int i = 1; i < n - 1; i ++ ){int tmp1 = abs(a[i] - a[i - 1]);int tmp2 = abs(a[i] - a[i + 1]);if (tmp1 < tmp2){g[i].push_back(make_pair(i - 1, 1));g[i].push_back(make_pair(i + 1, tmp2));}else{g[i].push_back(make_pair(i - 1, tmp1));g[i].push_back(make_pair(i + 1, 1));}}int m;cin >> m;vector<int> dist1(n, 1e9);dist1[0] = 0;priority_queue<PII, vector<PII>, greater<PII>> q;q.push(make_pair(0, 0));vector<bool> std(n);while (q.size()){auto t = q.top();q.pop();int ver = t.second, dd = t.first;if (std[ver]) continue;std[ver] = true;bool flag = false;for (int i = 0; i < g[ver].size(); i ++ ){int j = g[ver][i].first;if (dist1[j] > dist1[ver] + g[ver][i].second){dist1[j] = dist1[ver] + g[ver][i].second;q.push(make_pair(dist1[j], j));}}}vector<int> dist2(n, 1e9);dist2[n - 1] = 0;q.push(make_pair(0, n - 1));for (int i = 0; i < n; i ++ ) std[i] = false;while (q.size()){auto t = q.top();q.pop();int ver = t.second, dd = t.first;if (std[ver]) continue;std[ver] = true;bool flag = false;for (int i = 0; i < g[ver].size(); i ++ ){int j = g[ver][i].first;if (dist2[j] > dist2[ver] + g[ver][i].second){dist2[j] = dist2[ver] + g[ver][i].second;q.push(make_pair(dist2[j], j));}}}while (m -- ){int st, ed;cin >> st >> ed;st -- , ed -- ;if (dist1[ed] - dist1[st] < 0) cout << dist2[ed] - dist2[st] << '\n';else cout << dist1[ed] - dist1[st] << '\n';}
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1922D Berserk Monsters

题目链接

因为如果一个人的旁边两个人不改变的话，他在第一局没有挂，之后就都不会挂，所以我们先进行第一轮判断，之后就只需要判断挂的人旁边的两个人即可

#include <bits/stdc++.h>using namespace std;// #define int long longusing i64 = long long;typedef pair<int, int> PII;void solve()
{int n;cin >> n;vector<int> a(n + 1), d(n + 1);for (int i = 1; i <= n; i ++ ) cin >> a[i];for (int i = 1; i <= n; i ++ ) cin >> d[i];vector<int> pre(n + 1), nxt(n + 1);for (int i = 1; i <= n; i ++ ) pre[i] = i - 1;for (int i = 1; i <= n; i ++ ) nxt[i] = i + 1;nxt[n] = 0;vector<bool> st(n + 1);vector<int> die;vector<int> vt;for (int i = 1; i <= n; i ++ ) vt.push_back(i);for (int i = 0; i < n; i ++ ){int cnt = 0;for (auto t : vt){if (st[t]) continue;int tmp = d[t];// if (pre[t] && !st[pre[t]]) tmp -= a[pre[t]];// if (nxt[t] && !st[nxt[t]]) tmp -= a[nxt[t]];// if (tmp < 0)if (a[pre[t]] + a[nxt[t]] > d[t]){die.push_back(t);cnt ++ ;st[t] = true;}}vt.clear();for (auto t : die){if (nxt[t]) pre[nxt[t]] = pre[t];if (pre[t]) nxt[pre[t]] = nxt[t];if (!st[pre[t]] && pre[t]) vt.push_back(pre[t]);if (!st[nxt[t]] && nxt[t]) vt.push_back(nxt[t]);}die.clear();cout << cnt << ' ';}cout << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1805A We Need the Zero

题目链接

熟悉位运算的性质：偶数个相同的数xor后是0，奇数个相同的数xor后是本身

#include <bits/stdc++.h>using namespace std;#define int long longusing i64 = long long;
using i128 = __int128_t;typedef pair<int, int> PII;void solve()
{int n;cin >> n;vector<int> a(n);for (int i = 0; i < n; i ++ ) cin >> a[i];if (n % 2 == 0){int tmp = 0;for (int i = 0; i < n; i ++ ) tmp ^= a[i];if (tmp == 0) cout << 1 << '\n';else cout << -1 << '\n';}else{int tmp = 0;for (int i = 0; i < n; i ++ ) tmp ^= a[i];cout << tmp << '\n';}
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1805B The String Has a Target

题目链接

记录一下最小字符，挪到最前面就行

#include <bits/stdc++.h>using namespace std;#define int long longusing i64 = long long;
using i128 = __int128_t;typedef pair<int, int> PII;vector<i64> p(65);void solve()
{int n;cin >> n;string s;cin >> s;char minn = 'z' + 1;int pos = -1;for (int i = n - 1; i >= 0; i -- ){if (s[i] < minn){minn = s[i];pos = i;}}if (minn == s[0]){if (pos != 0){cout << minn;for (int i = 0; i < n; i ++ ){if (i != pos) cout << s[i];}}else cout << s;}else{cout << minn;for (int i = 0; i < n; i ++ ){if (i != pos) cout << s[i];}}cout << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1805C Place for a Selfie

题目链接

联立方程看有没有符合条件的解，没有什么坑点

#include <bits/stdc++.h>using namespace std;#define int long longusing i64 = long long;
using i128 = __int128_t;typedef pair<int, int> PII;vector<i64> p(65);void solve()
{int n, m;cin >> n >> m;vector<int> k(n);for (int i = 0; i < n; i ++ ) cin >> k[i];sort(k.begin(), k.end());while (m -- ){int a, b, c;cin >> a >> b >> c;if (c < 0){cout << "NO\n";continue;}double tmp1 = b - 2 * sqrt(a * c);double tmp2 = b + 2 * sqrt(a * c);int pos1 = upper_bound(k.begin(), k.end(), tmp1) - k.begin();int pos2 = lower_bound(k.begin(), k.end(), tmp2) - k.begin() - 1;if (pos1 < 0 && pos2 >= n) cout << "NO\n";else if (pos1 >= 0 && pos1 < n && k[pos1] > tmp1 && k[pos1] < tmp2) cout << "YES\n" << k[pos1] << '\n';else if (pos2 >= 0 && pos2 < n && k[pos2] > tmp1 && k[pos2] < tmp2) cout << "YES\n" << k[pos2] << '\n';else cout << "NO\n";}
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1805D A Wide, Wide Graph

题目链接

考点是树的直径，不太熟悉于是补题的时候去学了下，详细内容见另一篇博客

#include <bits/stdc++.h>using namespace std;#define int long longusing i64 = long long;
using i128 = __int128_t;typedef pair<int, int> PII;vector<i64> p(65);void solve()
{int n;cin >> n;vector<vector<int>> g(n + 1);for (int i = 1; i < n; i ++ ){int u, v;cin >> u >> v;g[u].push_back(v), g[v].push_back(u);}int p, q;int c = 0;vector<int> d(n + 1);function<void(int, int)> dfs = [&](int u, int fa){for (int i = 0; i < g[u].size(); i ++ ){int j = g[u][i];if (j == fa) continue;d[j] = d[u] + 1;if (d[j] > d[c]) c = j;dfs(j, u);}};dfs(1, 0);p = c, c = 0;d[p] = 0;dfs(p, 0);q = c;int zj = d[c];vector<int> dist1(n + 1, 0x3f3f3f3f);vector<int> dist2(n + 1, 0x3f3f3f3f);function<void(int, int)> bfs = [&](int start, int idx){if (idx == 1) dist1[start] = 0;if (idx == 2) dist2[start] = 0;priority_queue<PII, vector<PII>, greater<PII>> q;q.push(make_pair(0, start));vector<bool> st(n + 1);while (q.size()){auto t = q.top();q.pop();int dd = t.first, ver = t.second;if (st[ver]) continue;st[ver] = true;for (int i = 0; i < g[ver].size(); i ++ ){int j = g[ver][i];if (idx == 1 && (dist1[j] == -1 || dist1[j] > dist1[ver] + 1)){dist1[j] = dist1[ver] + 1;q.push(make_pair(dist1[j], j));}else if (idx == 2 && dist2[j] > dist2[ver] + 1){dist2[j] = dist2[ver] + 1;q.push(make_pair(dist2[j], j));}}}};bfs(p, 1);bfs(q, 2);vector<int> dist(n + 1);for (int i = 1; i <= n; i ++ ) dist[i] = max(dist1[i], dist2[i]);dist[p] = dist[q] = 0;sort(dist.begin(), dist.end());for (int i = 1; i <= n; i ++ ){int cnt = lower_bound(dist.begin(), dist.end(), i) - dist.begin();if (zj < i) cnt -- ;else cnt -= 2;cout << cnt << ' ';}cout << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}