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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
1 #include<stdio.h> 2 #include<algorithm> 3 #include<string.h> 4 using namespace std; 5 int dp[20000] ; 6 int main() { 7 // freopen("a.txt" , "r" , stdin ) ; 8 int w[4000] , D[4000] ; 9 int M ; 10 int n ; 11 12 int result = 0 ; 13 while ( scanf("%d%d" , &n , &M ) != EOF ) { 14 for (int i = 0 ; i < n ; i++ ) { 15 scanf("%d%d" , &w[i] , &D[i] ) ; 16 } 17 memset (dp , 0 , sizeof(dp) ) ; 18 for (int i = 0 ; i < n ; i++ ) { 19 for (int j = M ; j >= w[i] ; j-- ) { 20 dp[j] = max ( dp[j] , dp[j - w[i]] + D[i] ) ; 21 } 22 } 23 result = -9999999 ; 24 for (int i = 0 ; i <= M ; i++ ) { 25 if ( result < dp[i] ) 26 result = dp[i] ; 27 // printf ("%d " , dp[i] ) ; 28 } 29 30 printf("%d\n" , result ) ; 31 } 32 return 0 ; 33 }
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原文地址：http://www.cnblogs.com/get-an-AC-everyday/p/4269194.html
