1200: [HNOI2005]木梳

Description

  

Input

第一行为整数L，其中4<=L<=100000，且有50%的数据满足L<=104，表示木板下侧直线段的长。第二行为L个正整数A1,A2,…,AL，其中1

Output

仅包含一个整数D，表示为使梳子面积最大，需要从木板上挖掉的格子数。

Sample Input

9

4 4 6 5 4 2 3 3 5

Sample Output

3

 

太坑爹了，证明在这里

最终方案中的a'[i]只可能是a[j]+k{|i-j|<=2,|k|<=1}

然后直接dp就行了

var
    a:array[0..100010]of longint;
    f:array[0..100010,-2..2,-1..1,0..1]of int64;
    h:array[0..100010,-2..2,-1..1]of longint;
    n:longint;
    ans:int64;
 
function min(x,y:int64):int64;
begin
    if x<y then exit(x);
    exit(y);
end;
 
procedure init;
var
    i:longint;
begin
    read(n);
    for i:=1 to n do
      read(a[i]);
    ans:=1<<60;
    fillchar(f,sizeof(f),122);
end;
 
procedure work;
var
    i,j,k,j1,k1:longint;
begin
    for j:=0 to 2 do
      if 1+j<=n then
      for k:=-1 to 1 do
        if a[1+j]+k<=a[1] then
        begin
          h[1,j,k]:=a[1+j]+k;
          f[1,j,k,0]:=a[1]-h[1,j,k];
          f[1,j,k,1]:=a[1]-h[1,j,k];
        end;
    for i:=2 to n do
      for j:=-2 to 2 do
        if (i+j>0)and(i+j<=n) then
        for k:=-1 to 1 do
          if a[i+j]+k<=a[i] then
          begin
            h[i,j,k]:=a[i+j]+k;
            for j1:=-2 to 2 do
              for k1:=-1 to 1 do
                if h[i-1,j1,k1]>h[i,j,k] then f[i,j,k,0]:=min(f[i,j,k,0],f[i-1,j1,k1,1]+a[i]-h[i,j,k])
                else
                  if h[i-1,j1,k1]<h[i,j,k] then f[i,j,k,1]:=min(f[i,j,k,1],f[i-1,j1,k1,0]+a[i]-h[i,j,k])
                  else
                    begin
                      f[i,j,k,0]:=min(f[i,j,k,0],f[i-1,j1,k1,0]+a[i]-h[i,j,k]);
                      f[i,j,k,1]:=min(f[i,j,k,1],f[i-1,j1,k1,1]+a[i]-h[i,j,k]);
                    end;
          end;
    for j:=-2 to 0 do
      for k:=-1 to 1 do
        ans:=min(ans,min(f[n,j,k,0],f[n,j,k,1]));
    write(ans);
end;
 
begin
    init;
    work;
end.