来源：poj3624 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 题意概括： bessie在商店买手串，手串需要用珠子串起来，每个珠子的重量和价值都不一样，给出bessie能带的最大重量求出手串的最大价值。 解题思路： 01背包问题，状态转移方程：dp[j]=max(dp[j],dp[j-w[i]]+c[i]); 代码：
#include<stdio.h> #include<string.h> int max(int x,int y); int main() { int m,n,w[14030],c[14030],dp[14030]; while(scanf("%d%d",&m,&n)!=EOF) { memset(dp,0,sizeof(dp)); for(int i=0;i<m;i++) { scanf("%d%d",&w[i],&c[i]); } for(int i=0;i<m;i++)//表示物品编号 { for(int j=n;j>=w[i];j--)//当体积为j时，装不装当前物品 { dp[j]=max(dp[j],dp[j-w[i]]+c[i]); } } printf("%d\n",dp[n]); } } int max(int x,int y) { if(x>y)return x; return y; }