POJ-3624 Charm Bracelet

Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K       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 Wi(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

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思路：01背包。我就不多说了。都会做。

代码如下：

/****************************************/
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <stack>
 #include <queue>
 #include <vector>
 #include <map>
 #include <string>
 #include <iostream>
 using namespace std;
/****************************************/
int f[12890];

int main()
{
	int n, V, cost, worth;
	while(~scanf("%d%d", &n, &V)) {
		for(int i = 0; i < n; i++) {
			scanf("%d%d", &cost, &worth);
			for(int j = V; j >= cost; j--)
				f[j] = max(f[j], f[j-cost] + worth);
		}
		printf("%d\n", f[V]);
	}
	return 0;
}