poj-3624-Charm Bracelet0-1背包

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

4 6
1 4
2 6
3 12
2 7

Sample Output

23

这个题目是经典的0-1背包问题，借此学习0- 1背包

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。

用子问题定义状态：即f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是：f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}。可以压缩空间，f[v]=max{f[v],f[v-c[i]]+w[i]}。

代码：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>

using namespace std;
#define N 3410

int w[N], v[N];
int dp[N << 2];
int main () {
	int n, V;
	while (scanf("%d%d", &n, &V) != EOF) {
		for (int i = 1; i <= n; i ++) {
			scanf("%d%d", &w[i], &v[i]);
		}
		memset(dp, 0, sizeof(dp));
		for (int i = 1;i <= n; i ++) {
			for (int total_w = V; total_w >= w[i]; total_w --) {
				dp[total_w] = max(dp[total_w - w[i]] + v[i], dp[total_w]);
			}
		}
		printf("%d\n", dp[V]);
	}
	return 0 ;
}