poj--3624--Charm Bracelet（动态规划 水题）

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POJ - 3624 Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Submit Status 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 #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int dp[100100]; struct node { int w,val; }edge[100100]; int main() { int m,n; while(scanf("%d%d",&m,&n)!=EOF) { memset(dp,0,sizeof(dp)); for(int i=0;i<m;i++) scanf("%d%d",&edge[i].w,&edge[i].val); for(int i=0;i<m;i++) for(int j=n;j>=edge[i].w;j--) dp[j]=max(dp[j],dp[j-edge[i].w]+edge[i].val); printf("%d\n",dp ); } return 0; }

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