（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;}