poj3628 Bookshelf 2（0/背包）

题意：就是给出n和b，然后给出n个数，用这n个数中的某些，求出一个和，这个和是>=b的最小值，输出最小值与b的差。

分析：这道题很简单，是很明显的01背包问题，这里的n个物品，每个物品的重量为c[i]，价值为w[i]并且c[i]==w[i]，，容量为所有c[i]的和sum。只要在f[]中从头开始找，找到一个最小>=b的就是题目要的解



Description

FarmerJohnrecentlyboughtanotherbookshelfforthecowlibrary,buttheshelfisgettingfilledupquitequickly,andnowtheonlyavailablespaceisatthetop.

FJhasNcows(1≤
N≤20)eachwithsomeheightofHi(1≤Hi≤1,000,000-theseareverytallcows).Thebookshelfhasaheightof
B(1≤B≤S,whereSisthesumoftheheightsofallcows).

Toreachthetopofthebookshelf,oneormoreofthecowscanstandontopofeachotherinastack,sothattheirtotalheightisthesumofeachoftheirindividualheights.Thistotalheightmust
benolessthantheheightofthebookshelfinorderforthecowstoreachthetop.

Sinceatallerstackofcowsthannecessarycanbedangerous,yourjobistofindthesetofcowsthatproducesastackofthesmallestheightpossiblesuchthatthestackcanreachthebookshelf.
Yourprogramshouldprinttheminimal'excess'heightbetweentheoptimalstackofcowsandthebookshelf.

Input

*Line1:Twospace-separatedintegers:
NandB

*Lines2..N+1:Linei+1containsasingleinteger:Hi

Output

*Line1:Asingleintegerrepresentingthe(non-negative)differencebetweenthetotalheightoftheoptimalsetofcowsandtheheightoftheshelf.

SampleInput

516
3
1
3
5
6

SampleOutput

1

代码:

#include<iostream>
usingnamespacestd;
intf[1000050],c[10000];
intmain()
{
inti,j,n,v,sum;
while(scanf("%d%d",&n,&v)!=EOF)
{
memset(f,0,sizeof(f));
memset(c,0,sizeof(c));
sum=0;
for(i=1;i<=n;i++)
{
scanf("%d",&c[i]);
sum+=c[i];
}
for(i=1;i<=n;i++)
{
for(j=sum;j>=c[i];j--)
f[j]=max(f[j],f[j-c[i]]+c[i]);
}
for(i=1;i<=sum;i++)
{
if(f[i]>=v)
{
printf("%d\n",f[i]-v);
break;
}
}
}
return0;
}