bzoj 2011

决策单调性，对于一个1D/1D（状态是一维，转移也是一维）的DP，如果DP的决策具有单调性，那么就可以做到O(nlogn)的复杂度完成DP。

感谢《1D/1D  动态规划优化初步》的作者。

 /**************************************************************
     Problem: 2216
     User: idy002
     Language: C++
     Result: Accepted
     Time:4916 ms
     Memory:14476 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <cmath>
 #define N 500010
 using namespace std;

 struct Trid {
     int p, l, r;
     Trid(){}
     Trid( int p, int l, int r ):p(p),l(l),r(r){}
 };

 int n;
 int aa[N];
 int f[N], g[N], h[N];
 Trid stk[N]; int top;

 double calc( int j, int i ) {
     return aa[j]-aa[i]+sqrt(abs(i-j));
 }
 void dodp( int dp[N] ) {
     stk[top=] = Trid(,,n);
     for( int i=; i<n; i++ ) {
         if( calc(stk[top].p,n)>calc(i,n) ) continue;

         while( stk[top].l>=i &&
                 calc(stk[top].p,stk[top].l)<calc(i,stk[top].l) )
             top--;
         if( stk[top].r==i- ) {
             stk[++top] = Trid( i, i, n );
         } else {
             int lf = max( stk[top].l+, i );
             int rg = min( stk[top].r+, n );
             int p = stk[top].p;
             while( lf<rg ) {
                 int mid=(lf+rg)>>;
                 if( calc(p,mid) > calc(i,mid) ) lf=mid+;
                 else rg=mid;
             }
             stk[top].r = lf-;
             stk[++top] = Trid( i, lf, n );
         }
     }
     for( int i=; i<=top; i++ )
         for( int j=stk[i].l; j<=stk[i].r; j++ )
             dp[j] = stk[i].p;
 }
 int main() {
     scanf( "%d", &n );
     for( int i=; i<=n; i++ )
         scanf( "%d", aa+i );
     dodp(f);
     reverse( aa+, aa++n );
     dodp(g);
     reverse( aa+, aa++n );

     for( int i=; i<=n; i++ )
         g[i] = n+-g[i];
     reverse( g+, g++n );

     for( int i=; i<=n; i++ )
         if( calc(f[i],i)>calc(g[i],i) ) h[i]=f[i];
         else h[i]=g[i];
     for( int i=; i<=n; i++ )
         printf( "%lld\n", (long long)ceil(calc(h[i],i)) );
 }