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
不多说, 就是01背包模板题, 啥变化都没有
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 13000
using namespace std;
int n, m;
int w[MAXN], v[MAXN], dp[MAXN];
int main() {
while(~scanf("%d%d", &n, &m)) {
for (int i = 0; i < n; i++) {
scanf("%d%d", &w[i], &v[i]);
}
memset(dp, 0, sizeof dp);
for(int i = 0; i < n; i++) {
for(int j = m; j >= w[i]; j--) {
dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
}
}
printf("%d\n", dp[m]);
}
return 0;
}
