usaco training 4.1.2 Fence Rails 题解

时间:2024-08-11 22:36:56

Fence Rails题解

Burch, Kolstad, and Schrijvers

Farmer John is trying to erect a fence around part of his field. He has decided on the shape of the fence and has even already installed the posts, but he's having a problem with the rails. The local lumber store
has dropped off boards of varying lengths; Farmer John must create as many of the rails he needs from the supplied boards.

Of course, Farmer John can cut the boards, so a 9 foot board can be cut into a 5 foot rail and a 4 foot rail (or three 3 foot rails, etc.). Farmer John has an `ideal saw', so ignore the `kerf' (distance lost during
sawing); presume that perfect cuts can be made.

The lengths required for the rails might or might not include duplicates (e.g., a three foot rail and also another three foot rail might both be required). There is no need to manufacture more rails (or more of
any kind of rail) than called for the list of required rails.

PROGRAM NAME: fence8

INPUT FORMAT

Line 1: N (1 <= N <= 50), the number of boards
Line 2..N+1: N lines, each containing a single integer that represents the length of one supplied board
Line N+2: R (1 <= R <= 1023), the number of rails
Line N+3..N+R+1: R lines, each containing a single integer (1 <= ri <= 128) that represents the length of a single required fence rail

OUTPUT FORMAT

A single integer on a line that is the total number of fence rails that can be cut from the supplied boards. Of course, it might not be possible to cut all the possible rails from the given boards.

农民John准备建一个栅栏来围住他的牧场。他已经确定了栅栏的形状,但是他在木料方面有些问题。当地的杂货储存商扔给John一些木板,而John必须从这些木板中找出尽可能多所需的木料。

  当然,John可以切木板。因此,一个9英尺的木板可以切成一个5英尺和一个4英尺的木料 (当然也能切成3个3英尺的,等等)。John有一把梦幻之锯,因此他在切木料时,不会有木料的损失。

  所需要的木料规格都已经给定。你不必切出更多木料,那没有用。

目录

[显示

[编辑]格式

PROGRAM NAME: fence8

INPUT FORMAT:

(file fence8.in)

第1行: N (1 <= N <= 50), 表示提供的木板的数目

第2到第N+1行: 包括一个整数表示提供的一块木板的长度

第N+2行:一个整数R (1 <= R <= 1023)表示所需的木板数目

第N+3行到第N+R+1行:包括一个整数(1 <= ri <= 128)表示所需木料的长度。

OUTPUT FORMAT:

(file fence8.out)

只有一行,一个数字,表示能切出的最多的所需木料的数目。当然,并不是任何时候都能切出所有所需木料。

[编辑]SAMPLE
INPUT

4
30
40
50
25
10
15
16
17
18
19
20
21
25
24
30

[编辑]SAMPLE
OUTPUT

   7

[编辑]提示1(小心地利用它们!)

这是广度多维背包问题,所以我们必须考虑数据。给你的搜索范围有很大的下限,所以我们必须使用深搜加上迭代深化来限制树的边界。然而,直接的迭代深化会变得很慢,所以剪枝是必要的。

---------------------------------------------------分割线-----------------------------------------------------------

惭愧啊,这道搜索题做得焦头烂额,调了一个多小时,几度想放弃。还好总算成功了!

以前我在VJ上也做到过这道题,但当时一直TLE。

为了叙述方便,我们把原来的木头叫原木,后来切的叫新木。

-------------------------------------------一开始我的想法-----------------------------------------------------

首先,把原木和新木快排一遍。

每次二分枚举可以切的木块数,再去验证。

验证时,设枚举的木块数为k。由贪心思想可知,我们挑前k小的新木来切割。

事实证明,每次先用大的原木切割会更快,因为相对来说,它能分的个数更多。(想到以前NOIP2012的文化之旅,倒着搜比正着搜更快)

然后再加点小的剪枝,取新木的前缀和,如果当前试的原木可以切完所有的新木,就直接return真。

可是,到第四个点就TLE了!!!

----------------------------------------------求助互联网------------------------------------------------------

太强了!我看到NOCOW里有个题解(新木):对于切剩下的board(无法再切下rail),统计一下总和。如果这个值大于board长度的总和减去rail长度的总和,一定无解,可以剪枝。这个剪枝最关键。

加了这个剪枝之后,我就立马过了,且最慢的点也才0.11ms。

真心膜拜那位大牛!

代码:(红色为增加代码)

/*
ID:juan1973
LANG:C++
PROG:fence8
*/

#include<stdio.h>
#include<algorithm>
using namespace std;
int a[51],b[1024],b_sum[1024],n,m,i,ans,sum,sum2;
bool check(int num,int now,int q,int limit)
{
  if (num==0) return true;
  if (q>limit) return false;
  if (b[num]!=b[num+1]) now=n;
  for (int i=now;i>0;i--)
  {
    if (a[i]>=b_sum[num]) return true;
    if (a[i]>=b[num])
    {
      a[i]-=b[num];
      if (check(num-1,i,q+(a[i]<b[1])?a[i]:0,limit)) {a[i]+=b[num];return true;}
      a[i]+=b[num];
    }
  }
  return false;
}
int erfen(int l,int r)
{
  if (l==r) return l;
  int mid=(l+r)/2+1;
  if (check(mid,n,,sum-b_sum[mid])) return erfen(mid,r);
                                     return erfen(l,mid-1);
}
int main()
{
  //freopen("fence8.in","r",stdin);
  //freopen("fence8.out","w",stdout);
  scanf("%ld",&n);
  for (i=1;i<=n;i++) {scanf("%ld",&a[i]);sum+=a[i];}
  sum2=sum;
  sort(a+1,a+n+1);
  scanf("%ld",&m);
  for (i=1;i<=m;i++) scanf("%ld",&b[i]);
  sort(b+1,b+m+1);
  for (i=1;i<=m;i++) b_sum[i]=b_sum[i-1]+b[i];
  for (i=1;i<=m;i++) {sum2-=b[i];if (sum2<0) break;}
  ans=erfen(0,i-1);
  printf("%ld\n",ans);
  //scanf("%ld",&n);
  return 0;
}