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poj--3624--Charm Bracelet（动态规划水题）

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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

Source

USACO 2007 December Silver

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[100100];
struct node
{
	int w,val;
}edge[100100];
int main()
{
	int m,n;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		memset(dp,0,sizeof(dp));
		for(int i=0;i<m;i++)
		scanf("%d%d",&edge[i].w,&edge[i].val);
		for(int i=0;i<m;i++)
		for(int j=n;j>=edge[i].w;j--)
		dp[j]=max(dp[j],dp[j-edge[i].w]+edge[i].val);
		printf("%d
",dp[n]);
	}
	return 0;
}

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原文地址：https://www.cnblogs.com/playboy307/p/5273787.html

poj--3624--Charm Bracelet（动态规划 水题）

poj--3624--Charm Bracelet（动态规划水题）