【bzoj2216】[Poi2011]Lightning Conductor
2017-11-09 07:31
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Description
已知一个长度为n的序列a1,a2,…,an。
对于每个1<=i<=n,找到最小的非负整数p满足 对于任意的j, aj < = ai + p – sqrt(abs(i-j))
Input
第一行n,(1<=n<=500000)
下面每行一个整数,其中第i行是ai。(0<=ai<=1000000000)
Output
n行,第i行表示对于i,得到的p
Sample Input
6
5
3
2
4
2
4
Sample Output
2
3
5
3
5
4
题解
f[i]=a[j]+sqrt(i-j)-ai
对于i,最优决策点为k,对于i~n,比k小的点不可能再成为最优决策点,可以采用类似整体二分的形式dp
http://blog.csdn.net/u012288458/article/details/50536768
代码
已知一个长度为n的序列a1,a2,…,an。
对于每个1<=i<=n,找到最小的非负整数p满足 对于任意的j, aj < = ai + p – sqrt(abs(i-j))
Input
第一行n,(1<=n<=500000)
下面每行一个整数,其中第i行是ai。(0<=ai<=1000000000)
Output
n行,第i行表示对于i,得到的p
Sample Input
6
5
3
2
4
2
4
Sample Output
2
3
5
3
5
4
题解
f[i]=a[j]+sqrt(i-j)-ai
对于i,最优决策点为k,对于i~n,比k小的点不可能再成为最优决策点,可以采用类似整体二分的形式dp
http://blog.csdn.net/u012288458/article/details/50536768
代码
#include<bits/stdc++.h> #define pa pair<int,int> typedef long long ll; const int N=10000000; const int mod=1000000007; const double eps=0.00000001; using namespace std; inline int read() { int x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int f[500005],g[500005]; int a[500005],n; void solve1(int l,int r,int L,int R) { if (l>r) return; int mid=(l+r)>>1; int p=0;double mx=0.0; for (int i=L;i<=R&&i<=mid;i++) if ((double)a[i]+sqrt(mid-i)>=mx) p=i,mx=(double)a[i]+sqrt(mid-i); f[mid]=a[p]+ceil(sqrt(mid-p)); solve1(l,mid-1,L,p); solve1(mid+1,r,p,R); } void solve2(int l,int r,int L,int R) { if (l>r) return; int mid=(l+r)>>1; int p=0;double mx=0.0; for (int i=R;i>=L&&i>=mid;i--) if ((double)a[i]+sqrt(i-mid)>=mx) p=i,mx=(double)a[i]+sqrt(i-mid); g[mid]=a[p]+ceil(sqrt(p-mid)); solve2(l,mid-1,L,p); solve2(mid+1,r,p,R); } int main() { n=read();for (int i=1;i<=n;i++) a[i]=read(); solve1(1,n,1,n); solve2(1,n,1,n); for (int i=1;i<=n;i++) printf("%d\n",max(f[i],g[i])-a[i]); return 0; }
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