优化后的01背包模板

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40141		Accepted: 17439

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 W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

Output

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

Sample Input

Sample Output

23
二维优化为一维，因为将01背包逐行输出后，发现前面的值对后面有影响，因此从后往前处理

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define PI 3.141592653589793238462
#define INF 1000000000
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll dp[13000];
ll x,y;
ll n,v;
int main()
{
    scanf("%lld%lld",&n,&v);
    mem(dp);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld%lld",&x,&y);
        for(int j=v;j>=0;j--)
        {
            dp[j]=dp[j];
            if(j-x>=0) dp[j]=max(dp[j],dp[j-x]+y);
        }
    }
    printf("%lld
",dp[v]);
    return 0;
}

优化后的01背包模板

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