题目描述

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.

有N件物品和一个容量为V的背包。第i件物品的重量是c[i]，价值是w[i]。求解将哪些物品装入背包可使这些物品的重量总和不超过背包容量，且价值总和最大。

输入输出格式

输入格式：

* 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

第一行：物品个数N和背包大小M

第二行至第N+1行：第i个物品的重量C[i]和价值W[i]

输出格式：

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

输出一行最大价值。

输入输出样例

输入样例#1

4 6
1 4
2 6
3 12
2 7

输出样例#1

23

思路

01背包

#include <stdio.h>
#include <iostream>
using namespace std;
int w[4001],v[4001],dp[20001],n,m;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int i,j;
	cin>>n>>m;
	for(i=1;i<=n;i++)
	cin>>w[i]>>v[i];
	for(i=1;i<=n;i++)
	{
		for(j=m;j>=w[i];j--)
		{
			if(dp[j-w[i]]+v[i]>dp[j])
			{
				dp[j]=dp[j-w[i]]+v[i];
			}
		}
	}
	cout<<dp[m]<<endl;
	return 0;
}