FZOJ--2214--Knapsack problem(背包)

Problem 2214 Knapsack problem

Accept: 5    Submit: 8

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 42 21 14 101 2

Sample Output

15

Source

第六届福建省大学生程序设计竞赛-重现赛（感谢承办方华侨大学）

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f
struct node
{
	int u,v;
}num[10010];
int dp[10010];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int m,n;
		scanf("%d%d",&n,&m);
		memset(dp,INF,sizeof(dp));
		int V=0;
		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&num[i].u,&num[i].v);
			V+=num[i].v;
		}
		dp[0]=0;
		for(int i=0;i<n;i++)
		{
			for(int j=V;j>=num[i].v;j--)
			{
				dp[j]=min(dp[j],dp[j-num[i].v]+num[i].u);
			}
		}
		for(int j=V;j>=0;j--)
		{
			if(dp[j]<=m)
			{
				printf("%d\n",j);
				break;
			}
		}
	}
	return 0;
}