0-1 backpack

0-1 backpack

Problem:

There are N items and a backpack with a capacity of V. The value of item I is c [I], and the weight is w [I]. Solving which items are loaded into a backpack can make the total weight of these items not exceed the size of the backpack, and the total value is the largest.

This problem is characterized by: each item has only one item, and you can choose to put it or not. F [I] [j] indicates that the current size of the backpack is j, and the maximum value of 1-I items is selected.

There are two different ways to solve the problem of a backpack: 1. Find the optimal solution smaller than or equal to the capacity of the backpack, that is, it is not necessarily filled with a backpack; 2. The optimal solution should be "just filled with a backpack.

1. Find the optimal solution that is smaller than or equal to the capacity of the backpack, that is, not necessarily full of the backpack;

Set f [0.. N] [0. v] = 0

2. The optimal solution should be "just filled with a backpack.

Then f [0] [1. v] =-∞ ensures that the optimal solution is not obtained when the backpack is not satisfied (the greatest value)

State transition equation:

Explanation:

1. the current size of the backpack j is less than the weight of the I items w [I-1], I items cannot be loaded, all do not put I items

2. items on the I-th item can be put down, selected, put or not put

• If not, the problem is converted to "pre-i-1 items are placed in a backpack with capacity of j"
• If put, the problem is converted to "putting the items in the front I-1 into the backpack with the remaining capacity j-w [I-1 ", the greatest value that can be gained at this time is f [I-1] [j-w [I-1] plus the value gained by placing the I item c [I-1]

In the case of "others", the value of f [I] [j] is the maximum value in the preceding two cases.

Code:

1 # include <iostream> 2 using namespace std; 3 int f [100] [100]; 4 int Min =-999999; 5 int package (int n, int v, int w [], int c []) // n-item quantity, v-backpack capacity, w [] weight of each item, c [] value of the corresponding item 6 {7 if (f [n] [v]! = 0) // This subproblem has been calculated. 8 return f [n] [v]; 9 if (n = 0 | v <= 0) // question type 1: maximum Value: 10 return 0; 11 12 // if (n = 0) // question type 2: maximum value when a backpack is full, set f [0] [0] = 0, f [0] [1 .. v] =-∞, that is, when the backpack is not satisfied, there is no optimal solution 13 // {14 // if (v! = 0) return Min; 15 // return 0; 16 //} 17 18 if (v <w [n-1]) // if the weight of the n-th item is greater than the current size of the backpack, the n-th item is not loaded. The size of the backpack is still v, select 19 f [n] [v] = package (n-1, v, w, c) from the remaining n-1 items ); 20 else // the current size of the backpack can hold the nth item. You can choose to hold or not hold the nth item 21 {22 int a = package (n-1, v, w, c ); // if n items are not loaded, select from the remaining n-1 items. The size of the backpack is still v23 int B = package (n-1, v-w [n-1], w, c) + c [n-1]; // installs the nth item, and the current value is the value of the nth item c [n] Plus (n-1, v-w [n]) optimal Solution 24 f [n] [v] = a> B? A: B; // select the most valuable solution 25} 26 return f [n] [v]; 27} 28 29 int main () 30 {31 memset (f, 0, sizeof (f); // set the boundary condition 32 int n, v, w [100], c [100]; 33 cout <"input the number of items :"; 34 cin> n; 35 cout <"input the volume of items:"; 36 cin> v; 37 cout <"input the weight of items :"; 38 for (int I = 0; I <n; I ++) 39 cin> w [I]; 40 cout <"input the value of items :"; 41 for (int I = 0; I <n; I ++) 42 cin> c [I]; 43 cout <"the bigest value is :"; 44 cout <package (n, v, w, c) <endl; 45}View Code

Result:

Question 1: Find the optimal solution that is less than or equal to the capacity of the backpack, that is, not necessarily full of the backpack;

Question 2: the optimal solution is required when "just filled with a backpack.