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

╰半橙微兮° 2022-07-31 09:56 181阅读 0赞

**Problem C Knapsack problem**

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Accept: 81 Submit: 434

Time Limit: 3000 mSec Memory Limit : 32768 KB

## ![prodesc.gif][] 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).

## ![prodesc.gif][] 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.

## ![prodesc.gif][] Output ##

For each test case, output the maximum value.

## ![prodesc.gif][] Sample Input ##

5 15

12 4

2 2

1 1

4 10

1 2

## ![prodesc.gif][] Sample Output ##

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;
    }

[prodesc.gif]: http://acm.fzu.edu.cn/image/prodesc.gif