背包问题---DP

 Problem 2214 Knapsack problem

Accept: 412    Submit: 1650
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 4

2 2

1 1

4 10

1 2

 Sample Output

15

 

代码如下:

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/*     0-1背包问题：     因为重量太大，肯定不能按照原来的标准模型套用，开不了那么大的数组     但是总价值却小于5000;       根据价值构造dp方程,找到最小的重量能够得到这个价值     dp[100]=15,就是代表使得价值为100的最小重量为15; */     #include <iostream> #include<cstring> #include<cstdio> using namespace std; const int maxs = 600; int n,B; int w[maxs]; int v[maxs]; int dp[5005]; int main() {     freopen("in.txt","r",stdin);     int T;     cin>>T;     while(T--)     {         int sum = 0;         scanf("%d%d\n",&n,&B);         for(int i=1;i<=n;i++)             scanf("%d%d",&w[i],&v[i]),sum+=v[i];         memset(dp,0x3f,sizeof(dp));         dp[0]=0;         for(int i=n;i>=1;i--)             for(int j=sum;j>=v[i];j--)                 dp[j]=min(dp[j],dp[j-v[i]]+w[i]);         for(int i=sum;i>=0;i--)             if(dp[i]<=B)             {                 printf("%d\n",i);                 break;             }       }     return 0; }