Bzoj1044: [haoi2008] wooden sticks

1044: [haoi2008] Time Limit: 10 sec memory limit: 162 MB
Submit: 1580 solved: 567
[Submit] [Status] Description

There are n wooden sticks. The length of the I wooden sticks is Li, and N wooden sticks are connected in turn. There are n-1 connections in total. now you can cut down a maximum of M connections. After the cut, N wooden sticks are divided into many segments, which requires the minimum length of the maximum segment, how many cutting methods are output to minimize the length of the maximum segment. and the result mod 10007...

Input

The first line of the input file has 2 numbers n, m. The next n rows have a positive integer Li, indicating the length of the I stick.

Output

There are two output numbers. The first number is the minimum length of the segment with the largest total length, and the second number is the number of cutting methods to meet the conditions.

Sample input3 2
1
1
10
Sample output10 2

Two methods: (1) (1) (10) and (1) (10)
Data range
N <= 50000, 0 <= m <= min (n-1, 1000 ).
1 <= LI <= 1000.
Hint Source

Question:

Question 2 + greed can be obtained

In the second simple statement, the worst case is O (Mn ^ 2), but considering that each contact has pre [I], indicates that the maximum J value of S [I]-s [J] <= ANS is met.

Having f [I, j] = sigma (F [K, J-1]) (pre [I] <= k <= i-1), considering that pre [I] is a single addition, so we can use monotonous queues for optimization to O (Mn), but we can also skip queues,

Only one temporary variable is used to record the weights and

Let's talk about the following:

1. It is hard to write this question.

2. When writing a program, it should be easy to understand and error-prone, but should be short and beautiful. Don't think about how to reduce the code as soon as you come up, or you will get dizzy with yourself.

Code:

 1 const maxn=50000+100;p=10007; 2 var i,j,n,m,x,ans,cnt,l,r,mid,t,sum:longint; 3     q,pre,len,s:array[0..maxn] of longint; 4     f:array[0..maxn,0..1] of longint; 5 procedure init; 6  begin 7   readln(n,m); 8   for i:=1 to n do begin readln(len[i]);s[i]:=s[i-1]+len[i];end; 9   s[n+1]:=maxlongint;10  end;11 function test(x:longint):boolean;12  var i,j,sum:longint;13  begin14   sum:=0;j:=0;15   for i:=1 to n do16    if sum+len[i]<=x then inc(sum,len[i])17    else begin inc(j);sum:=len[i];end;18  exit(j<=m);19  end;20 procedure main;21  begin22   l:=0;r:=1000000000;23   while l<r do24    begin25      mid:=(l+r)>>1;26      if test(mid) then r:=mid else l:=mid+1;27    end;28   ans:=l;29   j:=0;30   for i:=1 to n do31    begin32    while s[i]-s[j]>ans do inc(j);33    pre[i]:=j;34    end;35   fillchar(f,sizeof(f),0);36   i:=1;while s[i]<=ans do begin f[i,0]:=1;inc(i);end;37   t:=0;cnt:=0;38   for i:=1 to m do39    begin40     t:=1-t;41     l:=0;r:=0;sum:=0;42     for j:=1 to n do43      begin44      while (l<r) and (q[l]<pre[j]) do begin sum:=(sum-f[q[l],1-t]+p) mod p;inc(l);end;45      f[j,t]:=sum;46      inc(r);q[r]:=j;sum:=(sum+f[q[r],1-t]) mod p;47      end;48     cnt:=(cnt+f[n,t]) mod p;49    end;50  writeln(ans,‘ ‘,cnt);51  end;52 begin53   assign(input,‘input.txt‘);assign(output,‘output.txt‘);54   reset(input);rewrite(output);55   init;56   main;57   close(input);close(output);58 end.  

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