(01背包 当容量特别大的时候) Knapsack problem （fzu 2214）

http://acm.fzu.edu.cn/problem.php?pid=2214

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

 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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <map>
#include <vector>
#include <queue>
using namespace std;
typedef long long LL;
#define N 2600000
#define met(a, b) memset(a, b, sizeof(a))
#define INF 0x3f3f3f3f

int v[], w[];
int dp[N];

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int i, j, n, B, Max=, Index;

        scanf("%d%d", &n, &B);

        for(i=; i<=n; i++)
        {
            scanf("%d%d", &w[i], &v[i]);
            Max += v[i];
        }

        for(i=; i<=Max; i++)
            dp[i] = INF;
        dp[] = ;
        for(i=; i<=n; i++)
        for(j=Max; j>=v[i]; j--)
        {
           dp[j] = min(dp[j], dp[j-v[i]]+w[i]);
        }

        for(i=; i<=Max; i++)
            if(dp[i]<=B) Index = i;

        printf("%d\n", Index);

    }
    return ;
}