POJ 3624 Charm Bracelet(01背包模板题)

题目链接

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 52318		Accepted: 21912

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

AC代码（模板题）

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = ;
int dp[N];
int s, n;//背包容积和物品数量

struct Thing
{
    int w;
    int v;
}list[];

void init()
{
    for (int i = ; i <= s; i++)dp[i] = ;
}

void package()
{
    for (int i = ; i < n; i++)
    {
        for (int j = s; j >= list[i].w; j--)
        {
            dp[j] = max(dp[j], dp[j - list[i].w] + list[i].v);
        }
    }
}

int main()
{
    scanf("%d%d", &n, &s);
    for (int i = ; i < n; i++)scanf("%d%d", &list[i].w, &list[i].v);
    init();
    package();
    printf("%d", dp[s]);
    return ;
}

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

struct T
{
    int w, v;
}t[];

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = ; i < n; i++)scanf("%d%d", &t[i].w, &t[i].v);
    int dp[];
    memset(dp, , sizeof(dp));
    for (int i = ; i < n; i++)
    {
        for (int j = m; j >= t[i].w; j--)dp[j] = max(dp[j - t[i].w] + t[i].v, dp[j]);
    }
    printf("%d\n", dp[m]);
    return ;
}

二刷