POJ 3624 Charm Bracelet
2016-08-06 15:23
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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
简单背包,但是…数据范围二维数组开不起,所以要用一维数组。
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
简单背包,但是…数据范围二维数组开不起,所以要用一维数组。
#include <iostream> #include <cstdio> using namespace std; struct DC { int w, d; }p[3450]; int dp[13300]; int main() { int n, m; scanf("%d%d", &n, &m); for(int i = 1; i <= n; i++) scanf("%d%d", &p[i].w, &p[i].d); for(int i = 1; i <= n; i++) for(int j = m; j >= p[i].w; j--) dp[j] = max(dp[j - p[i].w] + p[i].d, dp[j]); cout<<dp[m]; return 0; }
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