FZU - 2214 Knapsack problem 01背包逆思维

Knapsack problem

 

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


01背包变形。将体积与价值倒着存，将不超出背包体积求最大价值问题转化为不超出总价值求最小体积问题。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <cctype>
#define MAX 5005
using namespace std;
//const int maxn = ;
const int INF = 0x3f3f3f3f;
typedef long long ll;

int w[MAX],v[MAX],f[MAX];
int min(int x,int y){
    return x<y?x:y;
}
int main(void){
    int t,V,W,n,i,j;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&V);
        W=0;
        memset(f,INF,sizeof(f));
        for(i=1;i<=n;i++){
            scanf("%d%d",&w[i],&v[i]);
            W+=v[i];
            
        }
        f[0]=0;
        for(i=1;i<=n;i++){
            for(j=W;j>=v[i];j--){
                if(f[j-v[i]]+w[i]<=V){
                    f[j]=min(f[j],f[j-v[i]]+w[i]);
                }
            }
        }
        for(i=W;i>=0;i--){
            if(f[i]<INF){
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}