（poj3624）Charm Bracelet（01背包）

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

分析：01背包，用滚动数组（一维数组）存储最大值

可以直接输入，不用数组存储，节省内存，但是不适合打印路径时候 使用

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define mem(a,n) memset(a,n,sizeof(a))
typedef long long LL;
const int N=13000;
const int INF=0x3f3f3f3f;
int dp[N];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        mem(dp,0);
        int w,d;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&w,&d);
            for(int j=m; j>=0; j--)
                if(j>=w)
                {
                    dp[j]=max(dp[j],dp[j-w]+d);
                    printf("dp[%d]=%d\n",j,dp[j]);
                }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}

用数组存储

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=3500;
int dp[13000],w[maxn],v[maxn];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0; i<n; i++)
        scanf("%d%d",&w[i],&v[i]);
    for(int i=0; i<n; i++)
        for(int j=m; j>=w[i]; j--)
            dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
    printf("%d\n",dp[m]);
    return 0;
}