HDU3448Bag Problem题解背包

Bag Problem

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/131072 K (Java/Others)
Total Submission(s): 445    Accepted Submission(s): 132

Problem Description

0/1 bag problem should sound familiar to everybody. Every earth man knows it well. Here is a mutant: given the capacity of a bag, that is to say, the number of goods the bag can carry (has nothing to do with the volume of the goods), and the weight it can carry. Given the weight of all goods, write a program that can output, under the limit in the above statements, the highest weight.

 

 

Input

Input will consist of multiple test cases The first line will contain two integers n (n<=40) and m, indicating the number of goods and the weight it can carry. Then follows a number k, indicating the number of goods, k <=40. Then k line follows, indicating the weight of each goods The parameters that haven’t been mentioned specifically fall into the range of 1 to 1000000000.

 

 

Output

For each test case, you should output a single number indicating the highest weight that can be put in the bag.

 

 

Sample Input

  

   5 100
8
8 64 17 23 91 32 17 12
5 10
3
99 99 99
  

 

 

Sample Output

  

   99
0
  

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（2）——Host by BUPT

二维费用背包

V比较大，用三维DP超时

n和m比较小，用set和map水过

#include<cstdio> #include<set> #include<map> using namespace std; int main() { int n,m,V; while(scanf("%d%d%d",&m,&V,&n)==3) { int i,w[45],x; set<int, greater<int> > d,t; set<int, greater<int> >::iterator it; map<int,int> s; for(i=0;i<n;i++) scanf("%d",w+i); d.insert(0); t=d; for(i=0;i<n;i++) { it=d.begin(); while(it!=d.end()) { x=*it; if(x+w[i]<=V&&s[x]<m) { t.insert(x+w[i]); s[x+w[i]]=s[x]+1; } it++; } d=t; } printf("%d/n",*d.begin()); } }