POJ 3624 - Charm Bracelet(01背包)

Charm Bracelet
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 34002 Accepted: 15080
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

解题思路:
01背包裸题,注意01背包的初始化.

如果要求必须填满,那么除了dp[0] = 0,其余都等于-∞,表示除了什么也不装价值为0的情况,其余都是非法解.

如果不是要求必须填满,都为0即可.

AC代码

#include<stdio.h>#include<algorithm>const int maxn = 40000;using namespace std;int w[maxn];int c[maxn];int package[maxn];int main(){    int n;    int m;    scanf("%d%d",&n,&m);    for(int i = 0;i < n;i++)    scanf("%d%d",&c[i],&w[i]);    for(int i = 0;i < n;i++)    {        for(int j = m;j >= c[i];j--)            package[j] = max(package[j],package[j-c[i]]+w[i]);    }    printf("%d\n",package[m]);    return 0;}