CF 459D Pashmak and Parmida’s problem

题目链接

最近感觉对数据结构题的反应度提升了，这一题是上午看的但是当时旁边没电脑，傍晚这会儿可能思路就出现了点偏差，想到了树状数组+前缀，应该是能独立写出来的一道题

正着反着各跑一遍前缀和，记录下每个数是第几次出现，把反着跑的结果存到树状数组里，之后遍历原数组，答案每次加上当前数后方的出现次数比当前数小的个数

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 1e6 + 10;
const int mod = 998244353;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;int tr[N];int lowbit(int x)
{return x & -x;
}void add(int pos, int x)
{for (int i = pos; i < N; i += lowbit(i)) tr[i] += x;
}int query(int pos)
{int res = 0;for (int i = pos; i > 0; i -= lowbit(i)) res += tr[i];return res;
}void solve()
{int n, ans = 0;cin >> n;vector<int> a(n);for (int i = 0; i < n; i ++ ) cin >> a[i];map<int, int> mp;vector<int> pre(n), back(n);for (int i = 0; i < n; i ++ ){mp[a[i]] ++ ;pre[i] = mp[a[i]];}mp.clear();for (int i = n - 1; i >= 0; i -- ){mp[a[i]] ++ ;back[i] = mp[a[i]];}for (int i = 0; i < n; i ++ ) add(back[i], 1);for (int i = 0; i < n; i ++ ){add(back[i], -1);ans += query(pre[i] - 1);}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 1388C Uncle Bogdan and Country Happiness

题目链接

也是上午看的一道题，思路非常正确，但还是经不住晕递归，调了很久，赛时看不到错误数据估计要调更久，唉

先dfs算一下每个点有多少人经过，然后再dfs给每个点分配开心的人数，如果有一个点不能分配就no，否则yes

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 1e6 + 10;
const int mod = 998244353;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n, m;cin >> n >> m;vector<int> p(n + 1), h(n + 1);for (int i = 1; i <= n; i ++ ) cin >> p[i];for (int i = 1; i <= n; i ++ ) cin >> h[i];vector<vector<int>> g(n + 1);for (int i = 0; i < n - 1; i ++ ) {int x, y;cin >> x >> y;g[x].push_back(y);g[y].push_back(x);}vector<int> pass(n + 1); // 每个点经过多少人function<void(int, int)> dfs1 = [&](int u, int fa){for (int i = 0; i < g[u].size(); i ++ ){int j = g[u][i];if (j == fa) continue;dfs1(j, u);pass[u] += pass[j];}pass[u] += p[u];};dfs1(1, -1);function<bool(int, int, int)> dfs2 = [&](int u, int fa, int ha){for (int i = 0; i < g[u].size(); i ++ ){int j = g[u][i];if (j == fa) continue;if ((h[j] < 0 && h[j] >= -pass[j]) || (h[j] >= 0 && h[j] <= pass[j])){if ((h[j] + pass[j]) % 2 != 0) return false;if ((h[j] + pass[j]) / 2 > min(ha, pass[j]) || (h[j] + pass[j]) / 2 < 0) return false;if (!dfs2(j, u, (h[j] + pass[j]) / 2)) return false;ha -= (h[j] + pass[j]) / 2;}else return false;}return true;};if (h[1] < -m || h[1] > m || (h[1] + m) % 2 != 0) cout << "NO\n";else if (dfs2(1, -1, (h[1] + m) / 2)) cout << "YES\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 1525D Armchairs

题目链接

dp

看到数据范围就想到dp，甚至连状态表示都构造出来了还搞不出来转移方程，真是没脑子啊

dp[i][j] 表示前 i 个需要变动的位置挪到前 j 个空位需要的最小代价

转移方程： dp[i][j] = min(dp[i][j - 1], dp[i - 1][j - 1] + abs(a[i] - b[j]))

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 1e6 + 10;
const int mod = 998244353;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n;cin >> n;vector<int> a, b;a.push_back(0), b.push_back(0);for (int i = 1; i <= n; i ++ ){int x;cin >> x;if (x) a.push_back(i);else b.push_back(i);}vector<vector<int>> dp(n + 1, vector<int>(n + 1, INF));for (int i = 0; i < b.size(); i ++ ) dp[0][i] = 0;for (int i = 1; i < a.size(); i ++ ){for (int j = 1; j < b.size(); j ++ ){dp[i][j] = min(dp[i][j - 1], dp[i - 1][j - 1] + abs(a[i] - b[j]));}}cout << dp[a.size() - 1][b.size() - 1] << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}

CF 220B Little Elephant and Array

题目链接

莫队的板子修改一下，太久没碰到过莫队居然已经完全忘记这个东西了

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;
typedef pair<PII, int> PPI;const int N = 1e6 + 10;
const int mod = 998244353;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n, m;cin >> n >> m;int size = sqrt(n);int bnum = ceil((double)n / size); // 把序列分成了多少块vector<int> belong(1e6 + 10);for (int i = 1; i <= bnum; i ++ )for (int j = (i - 1) * size + 1; j <= i * size; j ++ ){belong[j] = i; // 标记每个元素属于哪个块}vector<int> a(n + 1);for (int i = 1; i <= n; i ++ ) cin >> a[i];vector<PPI> q(m);for (int i = 0; i < m; i ++ ){cin >> q[i].first.first >> q[i].first.second;q[i].second = i;}function<bool(PPI a, PPI b)> cmp = [&](PPI a, PPI b){if (belong[a.first.first] != belong[b.first.first]) return belong[a.first.first] < belong[b.first.first];else{if (belong[a.first.first] % 2 == 1) return belong[a.first.second] < belong[b.first.second];else return belong[a.first.second] > belong[b.first.second];}};sort(q.begin(), q.end(), cmp);int l = 1, r = 0, cnt = 0;unordered_map<int, int> mp;vector<int> ans(m);// 添加数function<void(int)> add = [&](int pos){mp[a[pos]] ++ ;if (mp[a[pos]] == a[pos]) cnt ++ ;if (mp[a[pos]] == a[pos] + 1) cnt -- ;};// 删除数function<void(int)> del = [&](int pos){if (mp[a[pos]] == a[pos]) cnt -- ;if (mp[a[pos]] == a[pos] + 1) cnt ++ ;mp[a[pos]] -- ;};for (int i = 0; i < m; i ++ ){int ql = q[i].first.first, qr = q[i].first.second;while (l < ql) del(l++); // 如左指针在查询区间左方，左指针向右移直到与查询区间左端点重合while (l > ql) add(--l); // 如左指针在查询区间左端点右方，左指针左移while (r < qr) add(++r); // 右指针在查询区间右端点左方，右指针右移while (r > qr) del(r--); // 否则左移ans[q[i].second] = cnt;}for (int i = 0; i < m; i ++ ) cout << ans[i] << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;// cin >> t;while (t -- ){solve();}
}