Bzoj 3156 Defense Readiness (slope DP)

3156: Defense Readiness Time Limit:10 SEC Memory limit:512 MB
submit:837 solved:395
[Submit] [Status] [Discuss] Description Input

The first behavior of an integer n indicates the total length of the front.

The second row of n integers, and the first integer represents the cost of placing the guard tower in position I ai.

Output

A total of one integer that represents the minimum front cost value.

Sample Input

10
2 3 1 5 4 5 6 3 1 2

Sample Output
18

HINT

1<=n<=10^6,1<=ai<=10^9

Source

katharon+ #1

Ideas

Slope optimization DP.

The only thing to note is that the data over-range in the process needs to be specifically stated.

for the gross slope dp both long <_<

"Code 1"

1#include <cstdio>2#include <iostream>3 using namespacestd;4 5typedefLong LongLL;6 Const intN = 1e6+Ten;7 structPoint {LL x, y;8 }q[n],now;9 intN,l,r,a[n];Ten  One LL Cross (Point a,point b,point c) { A     return(b.x-a.x) * (C.Y-A.Y)-(B.Y-A.Y) * (c.x-a.x); - } - voidReadint&x) { the     CharC=getchar (); while(!isdigit (c)) c=GetChar (); -x=0; while(IsDigit (c)) x=x*Ten+c-'0', c=GetChar (); - } - intMain () { +     //freopen ("in.in", "R", stdin); -     //freopen ("Out.out", "w", stdout); + read (n); A      for(intI=1; i<=n;i++) read (A[i]); at      for(intI=1; i<=n;i++) { -          while(L<r && q[l].y-q[l].x*i >= q[l+1].y-q[l+1].x*i) l++; -now.x=i; -now.y=q[l].y-(LL) q[l].x*i+ (LL) i*i+A[i]; -          while(L<r && Cross (q[r-1],q[r],now) <=0) r--; -q[++r]=Now ; in     } -printf"%lld", q[r].y-(LL) n (n+1)/2);//(LL) n (n+1)/2 to     return 0; +}

Folding 1

"Code 2"

1#include <cstdio>2#include <cstring>3#include <iostream>4 using namespacestd;5 6typedefLong LongLL;7 Const intN = 1e6+Ten;8 9 intN,l,r,a[n],q[n]; LL F[n]; Ten  One DoubleSlopintKintj) { A     return(Double) (f[j]-f[k]+ (LL) j* (j+1)/2-(LL) k* (k +1)/2)/(J-k); - } - voidReadint&x) { the     CharC=getchar (); while(!isdigit (c)) c=GetChar (); -x=0; while(IsDigit (c)) x=x*Ten+c-'0', c=GetChar (); - } - intMain () { +     //freopen ("in.in", "R", stdin); -     //freopen ("Out.out", "w", stdout); + read (n); A      for(intI=1; i<=n;i++) read (A[i]); at      for(intI=1; i<=n;i++) { -          while(L<r && Slop (q[l],q[l+1]) <i) l++; -         intt=Q[l]; -f[i]=f[t]+ (LL) i* (I-T) + (LL) t* (t+1)/2-(LL) i* (i+1)/2+A[i]; -          while(L<r && Slop (q[r-1],q[r]) >slop (q[r],i)) r--; -q[++r]=i; in     } -printf"%lld", F[n]); to     return 0; +}

Folding 2

Bzoj 3156 Defense Readiness (slope DP)