Charm Bracelet(POJ-3624)
Problem 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 theN (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weightWi (1 ≤ Wi ≤ 400), a 'desirability' factorDi (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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
題意:給出 n 個物品,背包容量為 m,第 i 個物品重 w[i],價值 d[i],求背包能裝入物品的最大價值。
思路:01背包模版。
Source Program
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<set> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 100001 #define MOD 123 #define E 1e-6 using namespace std; int c[N]; int w[N],d[N]; int main() {int n,m;while(scanf("%d%d",&n,&m)!=EOF){memset(c,0,sizeof(c));for(int i=1;i<=n;i++)scanf("%d%d",&w[i],&d[i]);for(int i=1;i<=n;i++)for(int j=m;j>=w[i];j--)c[j]=max(c[j],c[j-w[i]]+d[i]);printf("%d\n",c[m]);}return 0; }?
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