hdu 2602 Bone Collector 01背包
Bone CollectorTime Limit: 1 Sec Memory Limit: 256 MB题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=2602
Description
Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
Input
The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
Output
One integer per line representing the maximum of the total value (this number will be less than 231).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
HINT
题意
01背包裸题
题解：
01背包，滚动数组优化一下
代码：
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH;
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
/*
inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH+48);
puts("");
}
*/
//**************************************************************************************
inline ll read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH+48);
puts("");
}
struct node
{
int v,vl;
};
node a;
int dp;
int main()
{
int t=read();
for(int cas=1;cas<=t;cas++)
{
memset(a,0,sizeof(a));
memset(dp,0,sizeof(dp));
int n=read(),v=read();
for(int i=1;i<=n;i++)
a.v=read();
for(int i=1;i<=n;i++)
a.vl=read();
for(int i=1;i<=n;i++)
{
for(int j=v;j>=a.vl;j--)
{
dp=max(dp,dp.vl]+a.v);
}
}
cout<<dp<<endl;
}
}
文档来源：51CTO技术博客https://blog.51cto.com/u_15303184/3111732
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