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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
大意：在一共M大的背包里面放N件东西，计算这N件东西的最大价值，很基础的背包问题，即在第二个for循环中从背包的最大容量开始，因为背包可以有剩余，然后减少，用递归，dp[j] = max(dp[j],dp[j-w[i]]+v[i]);
#include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> using namespace std; const int maxn = 222222; int w[maxn],v[maxn],dp[maxn]; int main(){ int n,m; memset(dp,0,sizeof(dp)); scanf("%d%d",&n,&m); for(int i = 1; i <= n ; i++) scanf("%d%d",&w[i],&v[i]); for(int i = 1; i <= n ; i++){ for(int j = m; j >= w[i];j--){ dp[j] = max(dp[j],dp[j-w[i]]+v[i]); } } printf("%d",dp[m]); }
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原文地址：http://www.cnblogs.com/zero-begin/p/4313189.html
