POJ3624 Charm Bracelet

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 41544 Accepted: 18040

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 61 42 63 122 7

Sample Output

23

01背包 dp[i][j] = max{dp[i-1][j],dp[i-1][j-w[i]]+v[i]}

#include <iostram>#include <cstring>#include <cstdio>#include <cmath>#include <string>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <map>#include <set>using namespace std;const int maxn = 10001;int dp[maxn];int w[maxn];int v[maxn];int main() {    int n;    int m;        while(cin >> n >> m) {        for(int i = 1;i <= n;i++) {            cin >> w[i] >> v[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]]+v[i]);            }        }                cout << dp[m] << endl;    }        return 0;}