Backpack IV
unbounded knapsack problem (UKP)
重复选择+唯一排列+装满可能性总数
Description
Given n items with size nums[i]
which an integer array and all positive numbers, no duplicates. An integer target denotes the size of a backpack. Find the number of possible fill the backpack.
Each item may be chosen unlimited number of times
Example
Given candidate items[2,3,6,7]
and target7
,
A solution set is:
[7]
[2, 2, 3]
Analysis + Solution
完全背包问题
借鉴Backpack III中完全背包问题的处理方法,k * nums[i - 1] <= j
作为nums[i - 1]
取得个数的限制条件。
状态:dp[i][j]
- 前i
个元素,加起来能装满j
大小的方法个数
状态转移方程:dp[i][j] += dp[i - 1][j - k * nums[i - 1]; (k = 0, 1, ..., j / nums[i - 1])
初始化条件: dp[0][0] = 1
,相当于说0个元素装满0大小,这个方法有1个。并且dp[0][i] = 0
(i > 0, i < target + 1)
注意这里是unique的组合方式,也就是说对于题目中的例子[2,2,3]
和 [2,3,2]
是一样的,只能计入1个。
因此外层循环用nums[]
,确保每次元素的选取组合不会与之前计算的重复。
public class Solution {
/**
* @param nums: an integer array and all positive numbers, no duplicates
* @param target: An integer
* @return: An integer
*/
public int backPackIV(int[] nums, int target) {
int n = nums.length;
int[][] dp = new int[n + 1][target + 1];
dp[0][0] = 1;
for (int i = 0; i <= n; i++) {
dp[i][0] = 1;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= target; j++) {
dp[i][j] = dp[i - 1][j];
for (int k = 1; k * nums[i - 1] <= j; k++) {
dp[i][j] += dp[i - 1][j - k * nums[i - 1]];
}
}
}
return dp[n][target];
}
}
Or Similarly
public int backPackIV(int[] nums, int target) {
// Write your code here
int m = target;
int []A = nums;
int f[][] = new int[A.length + 1][m + 1];
f[0][0] = 1;
for (int i = 1; i <= A.length; i++) {
for (int j = 0; j <= m; j++) {
for (int k = 0; k * A[i-1] <= j; k++) {
f[i][j] += f[i-1][j-A[i-1]*k];
}
} // for j
} // for i
return f[A.length][target];
}
优化空间复杂度为一维数组
为了与2D版有统一性,这里外循环用[0, nums.length],因此内层循环在使用时要用nums[i - 1]
public int backPackIV(int[] nums, int target) {
int n = nums.length;
int[] dp = new int[target + 1];
dp[0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = nums[i - 1]; j <= target; j++) {
dp[j] += dp[j - nums[i - 1]];
}
}
return dp[target];
}
因为这里不再需要nums.length + 1,因此也可以直接用[0, nums.length - 1]作为外循环。
public class Solution {
public int backPackIV(int[] nums, int target) {
int n = nums.length;
int[] f = new int[target + 1];
f[0] = 1;
for (int i = 0; i < n; ++i) {
for (int j = nums[i]; j <= target; j++) {
f[j] += f[j - nums[i]];
}
}
return f[target];
}
}
Reference
Jiuzhang Backpack Tutorial: https://www.jiuzhang.com/tutorial/backpack/471
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