FZU - 2214 Knapsack problem（0 1 背包++）

旧城等待， 2022-05-26 01:36 115阅读 0赞

[FZU - 2214 Knapsack problem][]

### Accept: 837 Submit: 3249   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 ##

15 1512 42 21 14 101 2

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

这道题是0-1背包类的问题，但是不同的是这道题的w\[i\]非常大，最大为1e9,

emmm,先粘一下普通代码：

for(int i=0;i<n;i++)
    			for(int j=0;j<=W;j++)
    				if(j<w[i])	dp[i+1][j]=dp[i][j];
    				else		dp[i+1][j]=max(dp[i][j],dp[i][j-w[i]]+v[i]);
    		printf("%d\n",dp[n][W]);

或者：

for(int i=n-1;i>=0;i--)
    			for(int j=0;j<=W;j++)
    				if(j<w[i])	dp[i][j]=dp[i+1][j];
    				else		dp[i][j]=max(dp[i+1][j],dp[i+1][j-w[i]]+v[i]);
    			
    		printf("%d\n",dp[0][W]);

这上边的代码时间复杂度是O(nW)

如果做这一道题的话，肯定会超时的啊

**所以这一道题DP的对象是价值 v\[i\]**

// 这一道题  DP的对象是价值  v[i] 
    
    //#include<bits/stdc++.h>		//CE
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int MAXN=505;
    const int MAXV=5000+5;
    const int INF=0x3f3f3f3f;
    int dp[MAXN][MAXV];
    int n,b;
    int w[MAXN],v[MAXN];
    int main()
    {
    	int t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%d %d",&n,&b);
    		for(int i=0;i<n;i++)	scanf("%d %d",&w[i],&v[i]);
    		memset(dp,0x3f,sizeof(dp));
    		dp[0][0]=0;
    		for(int i=0;i<n;i++)
    		{
    			for(int j=0;j<=MAXV;j++)
    			{
    				if(j<v[i])	dp[i+1][j]=dp[i][j];
    				else		dp[i+1][j]=min(dp[i][j],dp[i][j-v[i]]+w[i]);
    			}
    		}
    		int res=0;
    		for(int i=0;i<=5000;i++)	if(dp[n][i]<=b)	res=i;//这里的 i<=5000 一定要这样 
    		printf("%d\n",res);		
    	}
    	return 0;
    }

参考白书，第61页

本人实力有限，如有错误请您指出，谢谢。

[FZU - 2214 Knapsack problem]: http://acm.fzu.edu.cn/problem.php?pid=2214
[prodesc.gif]: http://acm.fzu.edu.cn/image/prodesc.gif