Poj 3624 charm bracelet Dynamic Planning (01 backpack problem)

Charm bracelet

Time limit:1000 ms		Memory limit:65536 K
Total submissions:15203		Accepted:6950

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
N(1 ≤N≤ 3,402) Available charms. Each charmIIn the supplied list has a weight
Wi(1 ≤Wi≤ 400), a 'inclurability 'factor
Di(1 ≤Di≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more
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:NAndM
* Lines 2 ..N+ 1: LineI+ 1 describes charmIWith two space-separated integers:
WiAndDi

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

Source

Usaco 2007 December silver

 

// Poj3624 01 backpack problem # include <stdio. h> # include <string. h>/* Note that the array must be large, and then the input does not need to judge the EOF input: 4 61 42 63 122 7 dynamic planning table: 0 1 2 3 4 5 60 0 0 0 0 0 01 0 4 4 4 4 4 42 0 4 6 10 10 10 103 0 4 6 12 16 18 22 4 0 4 7 11 13 19 23 */int w [20000], d [20000], F [20000]; int main () {int n, m; int I, j; scanf ("% d", & N, & M); memset (W, 0, sizeof (w); memset (D, 0, sizeof (d); memset (F, 0, sizeof (f )); for (I = 1; I <= N; I ++) scanf ("% d", & W [I], & D [I]); for (I = 0; I <= m; I ++) f [I] = 0; for (I = 1; I <= N; I ++) for (j = m; j> = W [I]; j --) {If (F [J] <(F [J-W [I] + d [I]) f [J] = f [J-W [I] + d [I];} printf ("% d \ n", F [m]); Return 0 ;}