NOI 2.6 动态规划 7113: Charm Bracelet

题目来源：http://noi.openjudge.cn/ch0206/7113/

7113: Charm Bracelet

总时间限制: 1000ms  内存限制: 65536kB

描述

Bessie has gone to the mall'sjewelry store and spies a charm bracelet. Of course, she'd like to fill it withthe best charms possible from the N(1≤ N≤ 3,402) availablecharms. Each charm iinthe 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 whoseweight is no more than M(1≤ M≤ 12,880).

Given that weight limit as aconstraint and a list of the charms with their weights and desirability rating,deduce the maximum possible sum of ratings.

 

输入

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

输出

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

样例输入

4 6
1 4
2 6
3 12
2 7

样例输出

23

来源

USACO 2007 December Silver

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解题思路

0-1背包，dp[i][j] (i=0,1,…,n;j=0,1,…,m)表示只考虑前i个物品且背包重量为j时的最大价值。递推式：

dp[i][j] =max(dp[i-1][j], dp[i-1][j-w[i]]+d[i])

一维数组优化的时候要注意倒着从m开始遍历到w[i]避免覆盖问题。

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代码

// 0-1背包
// 一维动态规划优化版本

#include<iostream>
#include<fstream>
using namespace std;

int dp[12890] = {};

int main()
{
#ifndef ONLINE_JUDGE
	ifstream fin("0206_7113.txt");
	int n,m,w,d,i,j;
	fin >> n >> m;
	for (i=1; i<=n; i++)
	{
		fin >> w >> d;
		for (j=m; j>=w; j--)					// 一维优化的时候要倒着遍历，这样不会造成覆盖问题
		{
			dp[j] = max(dp[j], dp[j-w]+d);
		}
	}
	fin.close();
	cout << dp[m];
	return 0;
#endif
#ifdef ONLINE_JUDGE
	int n,m,w,d,i,j;
	cin >> n >> m;
	for (i=1; i<=n; i++)
	{
		cin >> w >> d;
		for (j=m; j>=w; j--)
		{
			dp[j] = max(dp[j], dp[j-w]+d);
		}
	}
	cout << dp[m];
#endif
}