POJ Problem 3624 Charm Bracelet 【01背包】

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34019		Accepted: 15088

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

Source

USACO 2007 December Silver

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef __int64 Int;
typedef long long LL;
const double ESP = 1e-5;
const double Pi = acos(-1.0);
const int MAXN = 40000 + 10;
const int INF = 0x3f3f3f3f;
int dp[MAXN], N, M, w[MAXN], d[MAXN];
int main() {
    while (scanf("%d%d", &N, &M) != EOF) {
        for (int i = 1; i <= N; i++) {
            scanf("%d%d", &w[i], &d[i]);
        }
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= N; i++) {
            for (int j = M; j >= w[i]; j--) {
                dp[j] = max(dp[j], dp[j - w[i]] + d[i]);
            }
        }
        printf("%d
", dp[M]);
    }
    return 0;
}