POJ3624 - Charm Bracelet - 动态规划之01背包

1.题目描述：

Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37185   Accepted: 16254 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 Source USACO 2007 December Silver

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2.题意概述：
就是01背包，重量w，价值d

4.AC代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 4040
#define maxn 13000
using namespace std;
int w
, d
, dp[maxn];

int main()
{
int n, m;
while (scanf("%d%d", &n, &m) != EOF)
{
memset(dp, 0, sizeof(dp));
for (int i = 0; i < n; i++)
scanf("%d%d", &w[i], &d[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]] + d[i]);
}
}
printf("%d\n", dp[m]);
}
return 0;
}