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决策单调。整体二分。
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> #define maxn 500500 using namespace std; int n,a[maxn],f[maxn],g[maxn]; void dp1(int left,int right,int l,int r) { if (left>right) return; int mid=left+right>>1; int pos;double mx=0; for (int i=l;i<=r && i<=mid;i++) { if ((double)a[i]+sqrt(mid-i)>mx) { mx=(double)a[i]+sqrt(mid-i); pos=i; } } f[mid]=a[pos]+ceil(sqrt(mid-pos)); dp1(left,mid-1,l,pos);dp1(mid+1,right,pos,r); } void dp2(int left,int right,int l,int r) { if (left>right) return; int mid=left+right>>1; int pos;double mx=0; for (int i=r;i>=l && i>=mid;i--) { if ((double)a[i]+sqrt(i-mid)>mx) { mx=(double)a[i]+sqrt(i-mid); pos=i; } } g[mid]=a[pos]+ceil(sqrt(pos-mid)); dp2(left,mid-1,l,pos);dp2(mid+1,right,pos,r); } int main() { scanf("%d",&n); for (int i=1;i<=n;i++) scanf("%d",&a[i]); dp1(1,n,1,n);dp2(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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原文地址:http://www.cnblogs.com/ziliuziliu/p/5893833.html