Codeforces Round 881 (Div

E. Tracking Segments

给定初始长度为n，且全为0的序列a，然后给出m个线段，如果一个线段中1的个数严格大于0的个数，那么该线段称为一个漂亮线段，现在给出q次操作，每次操作使得序列a中位置x上的0变为1，请你求出第一次使得所有线段中出现漂亮线段的询问

题解：二分答案

• 容易发现答案具有单调性，所以我们可以考虑二分答案
• 我们发现在check的时候只要提前预处理好1和0个数前缀和数组即可，然后即可\(O(m)\)检验合法性
• 分析其时间复杂度为：\(O(m\times logq)\)

#include <bits/stdc++.h>
#define Zeoy std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0)
#define all(x) (x).begin(), (x).end()
#define rson id << 1 | 1
#define lson id << 1
#define int long long
#define mpk make_pair
#define endl '\n'
using namespace std;
typedef unsigned long long ULL;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-9;
const int N = 2e5 + 10, M = 4e5 + 10;

int n, m, q;
int L[N], R[N];
int qry[N];

bool check(int mid)
{
    vector<int> a(n + 10), pre0(n + 10), pre1(n + 10);
    for (int i = 1; i <= mid; ++i)
        a[qry[i]] = 1;
    for (int i = 1; i <= n; ++i)
    {
        pre1[i] = pre1[i - 1] + (a[i] == 1);
        pre0[i] = pre0[i - 1] + (a[i] == 0);
    }
    for (int i = 1; i <= m; ++i)
    {
        if (pre1[R[i]] - pre1[L[i] - 1] > pre0[R[i]] - pre0[L[i] - 1])
            return true;
    }
    return false;
}

void solve()
{
    cin >> n >> m;
    for (int i = 1; i <= m; ++i)
        cin >> L[i] >> R[i];
    cin >> q;
    for (int i = 1; i <= q; ++i)
        cin >> qry[i];
    int l = 1, r = q;
    bool flag = false;
    while (l <= r)
    {
        int mid = l + r >> 1;
        if (check(mid))
        {
            r = mid - 1;
            flag = true;
        }
        else
            l = mid + 1;
    }
    if (flag)
        cout << l << endl;
    else
        cout << -1 << endl;
}
signed main(void)
{
    Zeoy;
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}

F1. Omsk Metro (simple version)

给一棵有根树，每次给定点 v，询问从根节点到 v 路径上的某个连续子段的和能否等于k

题解：\(dp\)——最大/最小子段和

• 容易发现从根节点到\(v\)的路径上子段和的取值范围为\([最小子段和，最大字段和]\)
• 所以我们可以\(dp\)即可
• 状态转移时有3种选择：
• 选择接在前面的子段上
• 不接在前面的字段上，变成一个新的子段
• 选择成为空子段
• 查询时验证\(k\)是否在范围内即可

#include <bits/stdc++.h>
#define Zeoy std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0)
#define all(x) (x).begin(), (x).end()
#define rson id << 1 | 1
#define lson id << 1
#define int long long
#define mpk make_pair
#define endl '\n'
using namespace std;
typedef unsigned long long ULL;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-9;
const int N = 2e5 + 10, M = 4e5 + 10;

int n;

void solve()
{
    cin >> n;
    int idx = 1;
    vector<int> fmin(n + 10, INF), fmax(n + 10, -INF);
    vector<int> mi(n + 10, INF), mx(n + 10, -INF);
    fmin[1] = 0;
    fmax[1] = 1;
    mi[1] = min({mi[1], fmin[1], 0LL});
    mx[1] = max({mx[1], fmax[1], 0LL});
    for (int i = 1; i <= n; ++i)
    {
        char op;
        int u, v, k, x;
        cin >> op;
        if (op == '+')
        {
            cin >> u >> x;
            idx++;
            fmin[idx] = min({fmin[u] + x, x, 0LL});
            fmax[idx] = max({fmax[u] + x, x, 0LL});
            mi[idx] = min(mi[u], fmin[idx]);
            mx[idx] = max(mx[u], fmax[idx]);
        }
        else
        {
            cin >> u >> v >> k;
            if (mi[v] <= k && mx[v] >= k)
                cout << "YES" << endl;
            else
                cout << "NO" << endl;
        }
    }
}
signed main(void)
{
    Zeoy;
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}