lc2940 找到Alice和Bob可以相遇的建筑

给出数组H[n]和多组询问Q[m]，其中Q[i]={a[i],b[i]}表示查询最靠左的下标j，使得a[i]和b[i]都可以移到j处。从x处能移到y处的前提是x<y并且H[x]<H[y]。
1<=n<=5e4; 1<=H[i]<=1e9; 1<=m<=5e4; 0<=a[i],b[i]<=n-1

相当于找最靠左的上限，可以用st表或线段树来维护区间最大值，然后二分找最左。由于是静态数据，这里选择st表来维护。另外要注意a[i]=b[i]的情况，这点在二分check里不好处理，因此在外层特判。

#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=b;i>=a;i--)

const int N = 50004;
int n, a[N], Min[N][21], Max[N][21];
int qmin(int l, int r) {
    int k = log2(r+1-l);
    return min(Min[l][k], Min[r+1-(1<<k)][k]);
}
int qmax(int l, int r) {
    int k = log2(r+1-l);
    return max(Max[l][k], Max[r+1-(1<<k)][k]);
}
void build() {
    rep(i,1,n) Min[i][0] = Max[i][0] = a[i];
    rep(j,1,20) {
        for (int i = 1; i+(1<<j) <= n+1; i++) {
            Min[i][j] = min(Min[i][j-1], Min[i+(1<<(j-1))][j-1]);
            Max[i][j] = max(Max[i][j-1], Max[i+(1<<(j-1))][j-1]);
        }
    }
}

class Solution {
public:
    vector<int> leftmostBuildingQueries(vector<int>& H, vector<vector<int>>& Q) {
        n = H.size();
        rep(i,1,n) a[i] = H[i-1];
        build();
        int m = Q.size();
        vector<int> ans(m);
        for (int i = 0; i < m; i++) {
            int A = Q[i][0]+1;
            int B = Q[i][1]+1;
            if (A > B) swap(A,B);
            if (A == B || a[A] < a[B]) {
                ans[i] = B;
                continue;
            }
            int lo = B, hi = n, mid;
            while (lo < hi) {
                mid = lo + (hi-lo) / 2;
                int q = qmax(lo,mid);
                if (q > a[A] && q >= a[B])
                    hi = mid;
                else
                    lo = mid+1;
            }
            if (B <= lo && lo <= n && qmax(lo,lo) > a[A] && qmax(lo,lo) >= a[B])
                ans[i] = lo;
            else
                ans[i] = 0;
        }
        for (auto &i : ans) i--;
        return ans;
    }
};