FOJ Problem 2214 Knapsack problem

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

这一题跟平时的背包其实是一样的， 就是背包的容量太大了，枚举背包会超时，改为枚举价值

AC代码：

# include <cstdio>
# include <iostream>
# include <algorithm>
using namespace std;
int dp[2][5010];
int w[510], v[510];
const int inf=2000000000;
int main(){
	int n, i, j, k, e, sum_value, t, W;
	cin>>t;
	for(i=1; i<=t; i++){
		cin>>n>>W;
		sum_value=0;
		for(j=1; j<=n; j++){
			cin>>w[j]>>v[j];
			sum_value=sum_value+v[j];
		}
		e=1;
		for(j=0; j<=1; j++){
			for(k=0; k<=5009; k++){
				dp[j][k]=inf;
			}
		}
		dp[0][0]=dp[1][0]=0;
		for(j=1; j<=n; j++){
			e=1-e;
			for(k=1; k<=sum_value; k++){
				dp[1-e][k]=dp[e][k];
				if(k>=v[j])
				dp[1-e][k]=min(dp[e][k], dp[e][k-v[j]]+w[j]);
			}
		}
		for(j=sum_value; j>=0; j--){
			if(dp[1-e][j]<=W){
				cout<<j<<endl;
				break;
			}
		}
	}
	return 0;
}