第六届福建省大学生程序设计竞赛 Problem C Knapsack problem【背包问题】

Problem C Knapsack problem

Accept: 81    Submit: 434

Time Limit: 3000 mSec    Memory Limit : 32768 KB

Problem Description

Given a set of n items, each with a weight w[i] and a value v[i], determine a way to choose the items into a knapsack so that the total weight is less than or equal to a given limit B and the total value is as large as possible. Find the maximum total value. (Note that each item can be only chosen once).

Input

The first line contains the integer T indicating to the number of test cases.

For each test case, the first line contains the integers n and B.

Following n lines provide the information of each item.

The i-th line contains the weight w[i] and the value v[i] of the i-th item respectively.

1 <= number of test cases <= 100

1 <= n <= 500

1 <= B, w[i] <= 1000000000

1 <= v[1]+v[2]+...+v[n] <= 5000

All the inputs are integers.

Output

For each test case, output the maximum value.

Sample Input

1

5 15

12 4

2 2

1 1

4 10

1 2

Sample Output

15

解题思路：

由于B的取值范围过大，普通背包无法解决。所以要利用价值作为限制条件，找到重量最小的。最后再次筛选比限定质量小的。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int w[510];
int v[510];
long long f[5000+10];
int main(){
	int t,sum,n,b;
	scanf("%d",&t);
	while(t--){
		sum=0;
		scanf("%d%d",&n,&b);
		for(int i=1;i<=n;i++){
			scanf("%d%d",&w[i],&v[i]);
			sum+=v[i];
		}
		memset(f,0x3f,sizeof(f));
		f[0]=0;
		for(int i=1;i<=n;i++){
			for(int j=sum;j>=v[i];j--){
				f[j]=min(f[j],f[j-v[i]]+w[i]);
			}
		}
		for(int i=sum;i>=0;i--){
			if(f[i]<=b){
				printf("%d\n",i);
				break;
			}
		}
	}
	return 0;
}