整体二分(POJ2104 静态第k大查询)
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#include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> #include <string> using namespace std; const int lim = 1e9; int ans[200000]; int c[200000]; struct query { int l, r; int k, cnt; int id; }q[200000], b[200000]; struct point { int x, num; bool operator < (const point & rhs) const { return x < rhs.x; } }a[2000000]; int n, qu; int lowbit(int x) { return x & (-x); } void add(int id, int x) { while (id <= n) { c[id] += x; id += lowbit(id); } } int getsum(int id) { int res = 0; while (id) { res += c[id]; id -= lowbit(id); } return res; } void solve(int L, int R, int l, int r) { if (L == R) { for (int i = l; i <= r; ++i) { ans[q[i].id] = L; } return; } int mid = L + R >> 1; int curl = l, curr = r; int ll = 1, rr = n + 1; while (ll < rr) { //二分找出第一个大于等于L的a数组的下标 (lower_bound) int m = ll + rr >> 1; if (a[m].x >= L) { rr = m; } else { ll = m + 1; } } for (int i = rr; i <= n && a[i].x <= mid; ++i) //树状数组维护 [L, mid] 之间的 a[i].x 的个数 add(a[i].num, 1); for (int i = l; i <= r; ++i) //q[i].cnt 表示在[q[i].l, q[i].r]之间的a[i].x的个数 q[i].cnt = getsum(q[i].r) - getsum(q[i].l - 1); for (int i = rr; i <= n && a[i].x <= mid; ++i) //更新完q[i].cnt 后 恢复树状数组 add(a[i].num, -1); for (int i = l; i <= r; ++i) { if (q[i].cnt >= q[i].k) { b[curl++] = q[i]; } else { q[i].k -= q[i].cnt; b[curr--] = q[i]; } } for (int i = l; i <= r; ++i) { q[i] = b[i]; } if(curl != l) solve(L, mid, l, curl - 1); if(curr != r) solve(mid + 1, R, curr + 1, r); } void test3() { scanf("%d%d", &n, &qu); for (int i = 1; i <= n; ++i) { scanf("%d", &a[i].x); a[i].num = i; } sort(a + 1, a + n + 1); for (int i = 1; i <= qu; ++i) { scanf("%d%d%d", &q[i].l, &q[i].r, &q[i].k); q[i].id = i; } solve(-lim, lim, 1, qu); for (int i = 1; i <= qu; ++i) printf("%d\n", ans[i]); } int main() { test3(); return 0; }
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