bzoj 2957 楼房重建

Description

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给定n座楼，初始高度为0，每次可以改变某栋楼的高度，求每次改变高度之后从原点可以看到几栋楼

Solution 1

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一个比较显然的做法是分块，假设块大小是S,分为L块，维护每块中斜率单调上升的序列

每次暴力修改复杂度为O(S)

每次询问时对每块序列中二分第一个大于之前斜率的位置即可，复杂度O(L∗logN)

显然S=N/S∗logN即S=NlogN−−−−−−√时最优

Solution 2

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其实我们还可以用线段树做，修改一个值其实是对它前面的楼没有任何影响的，我们可以用线段树维护一个最大值以及能看到的楼的个数

正常进行单调修改，同时维护答案，每次维护的时候其实就是左端答案加上右端大于左端最大值的部分

复杂度O(Nlog2N)

Code1(分块)

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#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 100005;const double eps = 1e-10;int cnt[80];double a[N], b[80][1300];inline int read(int &t) {    int f = 1;char c;    while (c = getchar(), c < '0' || c > '9') if (c == '-') f = -1;    t = c - '0';    while (c = getchar(), c >= '0' && c <= '9') t = t * 10 + c - '0';    t *= f;}int main() {    int n, m, x, y;    read(n), read(m);    int S = (int)sqrt(n * log(n) / log(2) + 0.5), L = n / S + (n % S ? 1 : 0);    while (m--) {        read(x), read(y);        a[x - 1] = (double)y / x;        int bl = (--x) / S;        cnt[bl] = 0;        double now = 0.0;        for (int i = bl * S; i < (bl + 1) * S && i < n; ++i)                if (a[i] > now + eps)   b[bl][cnt[bl]++] = a[i], now = a[i];        now = 0.0;        int ans = 0;        for (int i = 0; i < L; ++i) {            int l = 0, r = cnt[i] - 1;            int t;            while (l <= r) {                int mid = l + r >> 1;                if (b[i][mid] > now + eps)  t = mid, r = mid - 1;                else l = mid + 1;            }            if (b[i][cnt[i] - 1] > now + eps)   ans += cnt[i] - t;            now = max(now, b[i][cnt[i] - 1]);        }        printf("%d\n", ans);    }    return 0;}

Code2(线段树)

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#include <bits/stdc++.h>using namespace std;#define ls (rt << 1)#define rs (rt << 1 | 1)const int N = 100005;int cnt[N << 2];double mx[N << 2];inline int read(int &t) {    int f = 1;char c;    while (c = getchar(), c < '0' || c > '9') if (c == '-') f = -1;    t = c - '0';    while (c = getchar(), c >= '0' && c <= '9') t = t * 10 + c - '0';    t *= f;}int calc(int rt, int l, int r, double x) {    if (l == r) return mx[rt] > x;    int mid = l + r >> 1;    if (mx[ls] <= x)    return calc(rs, mid + 1, r, x);    else return cnt[rt] - cnt[ls] + calc(ls, l, mid, x);}void change(int rt, int l, int r, int p, double x) {    if (l == r) {        mx[rt] = x;        cnt[rt] = 1;        return;    }    int mid = l + r >> 1;    if (p <= mid)   change(ls, l, mid, p, x);    else change(rs, mid + 1, r, p, x);    mx[rt] = max(mx[ls], mx[rs]);    cnt[rt] = cnt[ls] + calc(rs, mid + 1, r, mx[ls]);}int main() {    int n, m, x, y;    read(n), read(m);    while (m--) {        read(x), read(y);        change(1, 1, n, x, (double) y / x);        printf("%d\n", cnt[1]);    }    return 0;}