動態規劃解決POJ 3624
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 weightWi (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
View Code
#include<iostream> using namespace std; int f[12881]; //f[k]表示:當背包的容量為k時的最大價值 int max(int a,int b) { return a>b?a:b; } int main() { int Num,TotalWeight,i,k,j,weight[12881],value[12881]; cin>>Num>>TotalWeight; f[0]=0; for(i=0;i<Num;i++) cin>>weight[i]>>value[i]; for(i=0;i<Num;i++) for(k=TotalWeight;k>=weight[i];k--) { f[k]=max(f[k],(f[k-weight[i]]+value[i])); //在max中的兩個參數f[k], 和f[k-weight[i]]+value[i]都是表示在背包容量為k時的最大價值 //f[k]是這個意思,就不用說了。 //而f[k-weight[i]]+value[i]也表示背包容量為k時的最大價值是為什么呢? //首先,f[k-weight[i]]表示的是背包容量為k-weight[i]的容量,也就是說f[k-weight[i]] //表示的是容量還差weiht[i]才到k的價值,+walue[i]恰好彌補了差的這個價值。所以…… //如果你對f[k]=max(f[k],(f[k-weight[i]]+value[i]));這句話不是很清楚, //下面的這句代碼會對你有幫助 //cout<<"i="<<i+1<<" f["<<k<<"]="<<f[k]<<endl; } cout<<f[TotalWeight]<<endl; return 0; }
posted on 2011-09-23 22:23 More study needed. 閱讀(226) 評論(0) 收藏 舉報

浙公網安備 33010602011771號