Charm Bracelet
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 33860 | Accepted: 15017 |
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.
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
* 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
题解：
题意是你可以从n个珠宝中戴重量不超过m的珠宝，每个珠宝有wi的重量和di的价值，你要选择能带来价值最大的珠宝。
其实就是01背包
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[50000];
struct qer
{
int weight, volue;
}q[5000];
int main()
{
int n, m;
cin>>n>>m;
for(int i = 1;i <= n;i ++)
scanf("%d%d", &q[i].weight, &q[i].volue);
for(int i = 1;i <= n;i ++)
{
for(int j = m;j >= q[i].weight;j --)
{
dp[j] = max(dp[j], dp[j - q[i].weight] + q[i].volue);
}
}
cout<<dp[m]<<endl;
return 0;
}