ACM_01 Backpack (modified version)

Backpack 2

Time Limit: 2000/1000ms (Java/Others)

Problem Description:

There are n items whose weight and value are respectively Wi and Vi. Now select items whose total amount is just W from these items, and find the maximum value of the sum of all solutions.

Input:

The input contains multiple sets of test cases, each of which starts with two-digit integers n and W (1<=n<=10000, 1<=W<=1000), followed by n lines, each with two-digit integers Wi, Vi (1<=Wi<=10000, 1<=Vi<=100).

Output:

The output is one row, the maximum value of the sum of the values ​​across all scenarios. If there is no just filled condition, output "-1".

Sample Input:

3 4
1 2
2 5
2 1
3 4
1 2
2 5
5 1

Sample Output:

6
-1 
Problem solving idea: This is a modified version of the 01 backpack. Combined with the question, if there is no situation where it is just filled to be -1, we can initialize all the elements of the dp array to -1, so that by directly outputting dp[W], we can know that the total amount selected from these items is exactly W, The state transition equation can only be used when the sum of its values ​​also reaches the maximum and is not -1. 
AC code:

1 #include<bits/stdc++.h>
 2  using  namespace std;
 3  int w[ 10005 ],v[ 10005 ],dp[ 10005 ];
 4  int main()
 5  {
 6      int n,W;
 7      while (cin> >n>> W){
 8          memset(dp,- 1 , sizeof (dp)); // all initialized to -1 
9          dp[ 0 ]= 0 ; // default initialized to 0 
10          for ( int i= 0;i<n;++ i)
 11              cin>>w[i]>> v[i];
 12          for ( int i= 0 ;i<n;++ i){
 13              for ( int j=W;j >=w[i];-- j){
 14                  if (dp[jw[i]]!=- 1 )   // If it is not -1, it can be compared and selected 
15                      dp[j]=max(dp[j ],dp[jw[i]]+ v[i]);
 16              }
 17          }
 18          cout<<dp[W]<<endl;   // Directly output the value when the total amount is W 
19      }
 20      return  0 ;
 21}