【POJ3624】Charm Bracelet(01背包)
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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
注意要把数组开到题意指定大小,没看题目瞎做导致RE;
#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int maxn = 12900;int F[maxn][2];int dp[maxn];int max( int a,int b ){ return a>b? a:b;}int main(){ int n,m; memset(F,0,sizeof(F)); while(~scanf("%d%d",&n,&m)) { for( int i=1 ; i<=n ; i++ ) { scanf("%d%d",&F[i][0],&F[i][1]); } memset(dp,0,sizeof(dp)); for( int i=1 ; i<=n ; i++ ) for( int j=m ; j>=F[i][0] ; j-- ) { dp[j] = max( dp[j] , dp[j-F[i][0]]+F[i][1] ); } printf("%d\n",dp[m]); } return 0;}
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