POJ 3624 Charm Bracelet (线性dp 0/1背包)

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24286		Accepted: 10952

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 theN (1 ≤
N ≤ 3,402) available charms. Each charm i in the supplied list has a weightWi (1 ≤
Wi ≤ 400), a 'desirability' factorDi (1 ≤
Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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 andM

* 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

题目链接 ：http://poj.org/problem?id=3624

题目大意 : n种珠宝，每个珠宝有个重量w[i]，价值d[i]，求在重量不超过m的情况下的最大价值，（每个珠宝只能买一次）

题目分析：裸的0/1背包，dp[i][j]表示将前i个珠宝恰放入容量为j的背包时的最大价值则

dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - w[i]] + d[i])，空间压缩后： (w[i] <= k <= m) dp[k] = max(dp[k], dp[k - w[i]] + d[i])

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int w[3403], v[3403];
int dp[12881];
int main()
{
    int n, m;
    while(scanf("%d %d", &n, &m) != EOF)
    {
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++)
            scanf("%d %d", &w[i], &v[i]);
        for(int i = 1; i <= n; i++)
            for(int tot = m; tot >= w[i]; tot--)
                dp[tot] = max(dp[tot], dp[tot - w[i]] + v[i]);
        printf("%d\n", dp[m]);
    }
}