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poj 3624 Charm Bracelet 背包DP
阅读量:7056 次
发布时间:2019-06-28

本文共 2407 字,大约阅读时间需要 8 分钟。

Charm Bracelet

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://poj.org/problem?id=3624

Description

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di

Output

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

4 6
1 4
2 6
3 12
2 7

 

Sample Output

23

 

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 
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
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-9int Num;char CH[20];//const int inf=0x7fffffff; //нчоч╢Сconst 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[Num--]+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[Num--]+48); puts("");}struct node{ int v,vl;};node a[maxn];int dp[maxn];int main(){ int n,v; while(scanf("%d%d",&n,&v)!=EOF) { memset(a,0,sizeof(a)); memset(dp,0,sizeof(dp)); //n=read(),v=re9ad(); for(int i=1;i<=n;i++) a[i].vl=read(),a[i].v=read(); for(int i=1;i<=n;i++) { for(int j=v;j>=a[i].vl;j--) { dp[j]=max(dp[j],dp[j-a[i].vl]+a[i].v); } } cout<
<

 

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