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
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思路 |
简单01背包进行套用模板即可: for i=1…..N for v=V….0 f[v]=max(f[v], f[v-c[i]]+w[i]; |
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源码 |
#include<stdio.h> #include<string.h> #include<iostream> using namespace std; int main() { int w[3500], d[3500], m, n, i, bag[15000], j; while(scanf("%d%d", &n, &m)!=EOF) { memset(bag, 0, sizeof(bag)); for(i=1; i<=n; i++) scanf("%d%d", &w[i], &d[i]); for(i=1; i<=n; i++) for(j=m; j>=w[i]; j--) { bag[j]=max(bag[j], bag[j-w[i]]+d[i]); } printf("%d\n", bag[m]); } } |