[loj2157]避雷针
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不难发现,问题即求$\forall 1\le i\le n,\max_{1\le j\le n}h_{j}+\sqrt{|i-j|}-h_{i}$
其中$h_{i}$是常数,并将$j$分为$<i$和$j>$两部分分别处理(以下以前者为例)
构造函数$g_{j}(x)=h_{j}+\sqrt{x-j}$,问题也即求$\forall 1\le i\le n,\max_{1\le j<i}g_{j}(i)$
考虑$g_{j_{1}}(x)$和$g_{j_{2}}(x)$(保证$j_{1}<j_{2}$)的大小关系,注意到
$$
\Delta g(x)=g_{j_{2}}(x)-g_{j_{1}}(x)=(h_{j_{2}}+\sqrt{x-j_{2}})-(h_{j_{1}}+\sqrt{x-j_{1}})\\
\Delta g'(x)=(\sqrt{x-j_{2}})'-(\sqrt{x-j_{1}})'=\frac{1}{2}(\frac{1}{\sqrt{x-j_{2}}}-\frac{1}{\sqrt{x-j_{1}}})>0
$$
换言之,两者在交点左侧$g_{j_{1}}(x)>g_{j_{2}}(x),$交点右侧$g_{j_{1}}(x)<g_{j_{2}}(x)$(交点数量不超过1)
特别的,若$\Delta g(j_{2})>0$则认为交点在$-\infty$,若$\Delta g(\infty)<0$则认为交点在$\infty$(均指$x$坐标)
记$P(j_{1},j_{2})$表示$g_{j_{1}}(x)$和$g_{j_{2}}(x)$交点($x$坐标),从左到右枚举$i$并维护此时的函数"凸包",具体如下——
插入:加入函数$g_{i}(x)$时,设队次尾、队尾函数分别为$g_{j_{1}}(x)$和$g_{j_{2}}(x)$,不断弹出队尾直至$P(j_{1},j_{2})<P(j_{2},i)$
此时,若$P(j_{2},i)\ne \infty$,则在队尾加入$i$
查询:查询$i$处最大值时,设队首、对次首元素分别为$g_{j_{1}}(x)$和$g_{j_{2}}(x)$,不断弹出队首直至$i<P(j_{1},j_{2})$
此时,函数$g_{j_{1}}(x)$即为取到最大值的函数,即答案为$g_{j_{1}}(i)$
(两种情况均需注意队列大小不足的情况)
关于如何求$P(j_{1},j_{2})$,特判$\Delta g(j_{2})>0$和$\Delta g(\infty)<0$的情况(即交点不存在),进而即解方程
$$
h_{j_{1}}+\sqrt{x-j_{1}}=h_{j_{2}}+\sqrt{x-j_{2}}\\
\sqrt{x-j_{1}}-\sqrt{x-j_{2}}=h_{j_{2}}-h_{j_{1}}\\
\frac{(x-j_{1})-(x-j_{2})}{\sqrt{x-j_{1}}+\sqrt{x-j_{2}}}=h_{j_{2}}-h_{j_{1}}\\
\sqrt{x-j_{1}}+\sqrt{x-j_{2}}=\frac{j_{2}-j_{1}}{h_{j_{2}}-h_{j_{1}}}\\
$$
将其与第2个和第4个式子联立,不难解得
$$
2\sqrt{x-j_{1}}=(h_{j_{2}}-h_{j_{1}})+\frac{j_{2}-j_{1}}{h_{j_{2}}-h_{j_{1}}}\\
x=\left(\frac{(h_{j_{2}}-h_{j_{1}})+\frac{j_{2}-j_{1}}{h_{j_{2}}-h_{j_{1}}}}{2}\right)^{2}+j_{1}
$$
综上,时间复杂度为$o(n)$,可以通过
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 500005 4 #define oo 0x3f3f3f3f 5 int n,h[N],q[N],ans[N]; 6 double sqr(double x){ 7 return x*x; 8 } 9 int get_val(int i,int j){ 10 return h[j]+(int)ceil(sqrt(i-j)); 11 } 12 double get_point(int j1,int j2){ 13 if (h[j1]+sqrt(j2-j1)<h[j2])return -oo; 14 if (h[j1]>=h[j2])return oo; 15 int dj=j2-j1,dh=h[j2]-h[j1]; 16 return sqr((dh+1.0*dj/dh)/2)+j1; 17 } 18 void calc(){ 19 int l=1,r=0; 20 for(int i=1;i<=n;i++){ 21 while ((l<r)&&(get_point(q[l],q[l+1])<=i))l++; 22 if (l<=r)ans[i]=max(ans[i],get_val(i,q[l])); 23 while ((l<r)&&(get_point(q[r],i)<=get_point(q[r-1],q[r])))r--; 24 if ((l>r)||(get_point(q[r],i)!=oo))q[++r]=i; 25 } 26 } 27 int main(){ 28 scanf("%d",&n); 29 for(int i=1;i<=n;i++)scanf("%d",&h[i]); 30 calc(),reverse(h+1,h+n+1),reverse(ans+1,ans+n+1); 31 calc(),reverse(h+1,h+n+1),reverse(ans+1,ans+n+1); 32 for(int i=1;i<=n;i++)ans[i]=max(ans[i]-h[i],0); 33 for(int i=1;i<=n;i++)printf("%d\n",ans[i]); 34 return 0; 35 }View Code
标签:le,frac,int,Delta,j1,sqrt,避雷针,loj2157 来源: https://www.cnblogs.com/PYWBKTDA/p/15897902.html