【BZOJ2957】楼房重建

题面

题解

这是一种线段树套路,$CJK$美其名曰”楼房重建版线段树”

先把每个点的高度转为斜率
我们用线段树维护区间的最长上升子序列
合并两个儿子的信息时分最大值大于或小于当前节点左儿子最大值考虑即可

时间复杂度$O(nlog^2n)$

代码

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#include<bits/stdc++.h>
#define RG register
#define clear(x, y) memset(x, y, sizeof(x));
using namespace std;

inline int read()
{
int data=0, w=1;
char ch=getchar();
while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
if(ch=='-') w=-1, ch=getchar();
while(ch>='0'&&ch<='9') data=(data<<3)+(data<<1)+(ch^48), ch=getchar();
return data*w;
}

const int maxn(100010);
int n, m, tree[maxn << 2];
double val[maxn << 2];

#define son(i) (root << 1 | i)
inline int calc(double v, int root, int l, int r)
{
if(l == r) return val[root] > v;
int mid(l + r >> 1);
if(val[son(0)] <= v) return calc(v, son(1), mid + 1, r);
return tree[root] - tree[son(0)] + calc(v, son(0), l, mid);
}

inline void modify(int pos, double v, int root = 1, int l = 1, int r = n)
{
if(l == r) { tree[root] = 1, val[root] = v; return; }
int mid(l + r >> 1);
if(pos <= mid) modify(pos, v, son(0), l, mid);
else modify(pos, v, son(1), mid + 1, r);
val[root] = max(val[son(0)], val[son(1)]);
tree[root] = tree[son(0)] + calc(val[son(0)], son(1), mid + 1, r);
}

int main()
{
n = read(); m = read();
while(m--)
{
int x = read(), y = read();
modify(x, (double)y / x);
printf("%d\n", tree[1]);
}
return 0;
}
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