《动态规划》—POJ3624 CharmBracelet(01背包)

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 38399 Accepted: 16661

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 weightWi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 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: 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

 

题目大意：魔力手镯,      贝茜只能支持一个迷人的手镯  重量不超过M(1≤≤12880)，每一个魔力手镯的Wi,,Di.典型的01背包问题。背包容量为M.注意数组大小。

状态转移方程：dp[j] = max(dp[j] ,dp[j-vol[i]] + val[i])

 

#include <iostream> #include <cstdio> using namespace std; #define max(a,b) ((a)>(b)?(a):(b)) const int maxn = 3405; int main() { int n, v; int dp[4*maxn], val[maxn], vol[maxn]; while (scanf("%d %d", &n, &v) != EOF) { int i, j; for (i=0; i<n; i++) scanf("%d %d", &vol[i], &val[i]); memset(dp, 0, sizeof(dp)); for (i=0; i<n; i++) for (j=v; j>=vol[i]; j--) if (dp[j] < dp[j-vol[i]] + val[i]) //状态转移方程 dp[j] = dp[j-vol[i]] + val[i]; printf("%d\n", dp[v]); } return 0; }