2200左右的DS题单

zhujio / 2024-01-09 / 原文

Beautiful Subarrays

一眼转换为前缀和形式，然后字典树计数即可

#include <bits/stdc++.h> 
using namespace std;
#define endl "\n" 
typedef long long ll;

const int N = 1e6 + 100;
int Trie[N * 32][2], cnt, n, k;
int num[N * 32];
ll ans;

void insert(int x) {
    int now = 0;
    for (int i = 30; i >= 0; i--) {
        int j = x >> i & 1;
        if (!Trie[now][j]) Trie[now][j] = ++cnt;
        now = Trie[now][j];
        num[now]++;
    }
}

void query(int x) {
    int now = 0;
    for (int i = 30; i >= 0; i--) {
        int j = x >> i & 1;
        if (k >> i & 1) {
            if(!Trie[now][!j]) return;
            now = Trie[now][!j];
        } else {
            ans += num[Trie[now][!j]];
            if(!Trie[now][j]) return;
            now = Trie[now][j];
        }
    }
    ans += num[now];
}

void solve() {
    cin >> n >> k;
    ll now = 0;
    insert(0);
    for(int i = 1; i <= n; i++) {
        int x; cin >> x;
        now ^= x;
        insert(now);
        query(now);
    }
    cout << ans << endl;
}

signed main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

    // int T = 1; cin >> T;
    // while (T--) solve();
    solve();

    return 0;
}

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Closest Equals

类似于 HH的项链的离线做法

首先我们可以知道对于一个点计算距离真正有用的一定是前一个点（因为要求最短距离）

那么我们可以用线段树维护一个 $i-pre_i$ 最小值

为了保证在区间 [ l, r ] 之间我们离线下来每个询问，从 1 到 n 的处理每个询问

每到一个点 i 我们修改 $pre_i$ 为 $i-pre_i$ 这样保证在查询时只会查询到区间在 [ l , r ] 区间的最小值

#include <bits/stdc++.h> 
using namespace std;
#define endl "\n" 
typedef long long ll;

const int N = 5e5 + 100;

int a[N], pre[N], pos[N], Seg[N << 2], res[N];
vector<int> vec;
vector<array<int, 2>> g[N];

int get_idx(int x) {
    return lower_bound(vec.begin(), vec.end(), x) - vec.begin() + 1;
}

void pushup(int id) {
    Seg[id] = min(Seg[id << 1], Seg[id << 1 | 1]);
}

void modify(int id, int l, int r, int pos, int v) {
    if (l == r) {
        Seg[id] = v;
        return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid) modify(id << 1, l, mid, pos, v);
    else modify(id << 1 | 1, mid + 1, r, pos, v);
    pushup(id);
}

int query(int id, int l, int r, int x, int y) {
    if (x <= l && y >= r) return Seg[id];
    int mid = (l + r) >> 1, ans = INT_MAX;
    if (x <= mid) ans = min(ans, query(id << 1, l, mid, x, y));
    if(y > mid) ans = min(ans, query(id << 1 | 1, mid + 1, r, x, y));
    return ans;
}

void solve() {
    int n, m; cin >> n >> m;
    for (int i = 1; i <= 4 * N; i++) Seg[i] = INT_MAX;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        vec.push_back(a[i]);
    }

    for (int i = 1; i <= m; i++) {
        int l, r; cin >> l >> r;
        g[r].push_back({l, i});
    }
    sort(vec.begin(),vec.end());
    vec.erase(unique(vec.begin(), vec.end()), vec.end());

    for(int i = 1; i <= n; i++) {
        int id = get_idx(a[i]);
        pre[i] = pos[id];
        pos[id] = i;
        // cout << i << ' ' << pre[i] << endl; 
    }

    for(int r = 1; r <= n; r++) {
        if(pre[r]) modify(1, 1, n, pre[r], r - pre[r]);
        for(auto [l, id] : g[r]) {
            // cout << l << ' ' << r << endl;
            res[id] = query(1, 1, n, l, r);
        }
    } 

    for(int i = 1; i <= m; i++) {
        // cout << res[i] << endl;
        cout << (res[i] == INT_MAX ? -1 : res[i]) << endl;
    }
}

signed main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

    // int T = 1; cin >> T;
    // while (T--) solve();
    solve();

    return 0;
}

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Wine Factory (Easy Version)

用线段树维护一个类似于区间合并的 dp

参考 : Codeforces Hello 2024 ABCDF1F2 讲解

把每个区间多出来的水，多出来的体积，答案都维护出来

转移时一定是从左边多出来的水转移到右边多出来的体积

#include<bits/stdc++.h>
using namespace std;
#define endl "\n"
#define int long long 
typedef long long ll;

const int N = 5e5 + 100;

int a[N], b[N], c[N];

struct node {
    int s, r, d;
} Seg[N << 2];

void pushup(int id) {
    int v = min(Seg[id << 1].s, Seg[id << 1 | 1].r);
    Seg[id].d = Seg[id << 1].d + Seg[id << 1 | 1].d + v;
    Seg[id].s = Seg[id << 1].s + Seg[id << 1 | 1].s - v;
    Seg[id].r = Seg[id << 1].r + Seg[id << 1 | 1].r - v;

}

void modify(int id, int l, int r, int pos, int v1, int v2) {
    if(l == r) {
        Seg[id].d = min(v1, v2);
        Seg[id].s = max(0ll, v1 - v2);
        Seg[id].r = max(0ll, v2 - v1);
        return;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid) modify(id << 1, l, mid, pos, v1, v2);
    else modify(id << 1 | 1, mid + 1, r, pos, v1, v2);
    pushup(id);
}

void solve() {
    int n, q; cin >> n >> q;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) cin >> b[i];
    for (int i = 1; i < n; i++) cin >> c[i];

    for (int i = 1; i <= n; i++) modify(1, 1, n, i, a[i], b[i]);

    while (q--) {
        int p, x, y, z; cin >> p >> x >> y >> z;
        modify(1, 1, n, p, x, y);
        cout << Seg[1].d << endl;
    }
}

signed main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    
    // int T = 1; cin >> T;
    // while (T--) solve();
    solve();
    return 0;
}

View Code

Pillars

权值线段树（平衡树）优化 dp

朴素dp思路就是

$\large f[i]=max(f[j])+1$

满足$|\space b_j-b_i >=d|$

找 $\Large max$ 可以用权值线段树优化复杂度变成 nlogn

#include <bits/stdc++.h> 
using namespace std;
#define endl "\n" 
#define int long long
typedef long long ll;

const int N = 1e5 + 100;

int a[N], f[N], pre[N];
vector<ll> vec;
pair<int, int> Seg[N << 4];

int get_idx(int x) {
    return lower_bound(vec.begin(), vec.end(), x) - vec.begin() + 1;
}

pair<int, int> operator + (const pair<int, int> &l, const pair<int, int> &r) {
    pair<int, int> ans;
    auto [v1, pos1] = l;
    auto [v2, pos2] = r;
    if (v1 > v2) ans = l;
    else ans = r;
    return ans;
}

void pushup(int id) {
    Seg[id] = Seg[id << 1] + Seg[id << 1 | 1];
}

void modify(int id, int l, int r, int pos, int v) {
    if (l == r) {
        Seg[id] = {v, pos};
        return;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid) modify(id << 1, l, mid, pos, v);
    else modify(id << 1 | 1, mid + 1, r, pos, v);
    pushup(id);
}

pair<int, int> query(int id, int l, int r, int x, int y) {
    if (x <= l && y >= r) return Seg[id];
    int mid = (l + r) >> 1;
    pair<int, int> tl, tr;
    tl = tr = {INT_MIN, 0};
    if(x <= mid) tl = query(id << 1, l, mid, x, y);
    if(y > mid) tr = query(id << 1 | 1, mid + 1, r, x, y);
    return tl + tr;
}

void solve() {
    int n, d; cin >> n >> d;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        vec.push_back(a[i]);
        vec.push_back(a[i] + d);
        vec.push_back(a[i] - d);
    }
    vec.push_back(0); vec.push_back(1e15 + 100);
    sort(vec.begin(),vec.end());
    vec.erase(unique(vec.begin(), vec.end()), vec.end());   

    int sz = vec.size(), maxn = 0, mpos = -1;
    map<ll, int> vis;

    for (int i = 1; i <= n; i++) {
        int idl = get_idx(a[i] - d);
        int idr = get_idx(a[i] + d);
        int id = get_idx(a[i]);
        pair<int, int> l = query(1, 1, sz, 1, idl);
        pair<int, int> r = query(1, 1, sz, idr, sz);
        pair<int, int> ans = l + r;
        auto [v, pos] = ans;
        f[i] = v + 1;
        pre[i] = vis[pos];
        vis[id] = i; 
        if(f[i] > maxn) maxn = f[i], mpos = i;
        modify(1, 1, sz, id, f[i]);
    }

    // for(int i = 1; i <= n; i++) cout << i << ' ' << pre[i] << endl;
    vector<int> ans;
    cout << maxn << endl;
    ans.push_back(mpos);
    while (pre[mpos]) {
        mpos = pre[mpos];
        ans.push_back(mpos);
    }
    reverse(ans.begin(), ans.end());
    for(auto i : ans) cout << i << ' ';
    cout << endl;
}

signed main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

    // int T = 1; cin >> T;
    // while (T--) solve();
    solve();

    return 0;
}