Introduction to dynamic planning (DP) ------- longest non-downsample sequence (O (n ^ 2), dp -------

Introduction to dynamic planning (DP) ------- longest non-downsample sequence (O (n ^ 2), dp -------

When I first wrote a blog, I shared the longest non-Downloading subsequence of the DP problem I just learned, that is, the dynamic planning problem.

Problem description: You can enter a number or a number to output the maximum length of the non-descending subsequence. For example, 5, 3, 4, 8, 6, and 7, the longest non-descending subsequence length is 4.

In the face of such a problem, we must first define a "state" to represent its subproblem and find its solution. Note: In most cases, a State is only related to the State that appears before it, but independent of the State that follows it (a Way of Thinking ).

Suppose we want to calculate A [1], A [2],…, The maximum length of the non-descending subsequence of A [I], where I <N, then the above problem becomes A subproblem of the original problem (The problem scale is reduced, you can let I = 1, 2, 3, etc. for analysis) Then we define d (I), which indicates the length of the longest non-descending subsequence ending with A [I] In the first I number. OK. You can estimate that d (I) is the status we are looking for against the simple question in "Getting started. If we calculate all the values from d (1) to d (N), the answer we are looking for is the largest one. The status is found, and the next step is to find the state transition equation.

• The LIS length of the first number d (1) = 1 (sequence: 5)
• The LIS length of the first two numbers d (2) = 1 (sequence: 3; 3 front is not smaller than 3)
• The LIS length of the first three numbers d (3) = 2 (sequence: 3, 4; 4 there is a 3 smaller than it, So d (3) = d (2) + 1)
• The length of LIS in the first four numbers d (4) = 3 (sequence: 3, 4, 8; 3 before 8 is smaller than it, So d (4) = max {d (1), d (2), d (3)} + 1 = 3)

 

I can find it.D (I) = max {1, d (j) + 1}, where j <I, A [j] <= A [I]

 

The following is the code (c ):

1 # include <stdio. h> 2 # define Max 10 3 int list (int A [], int n); 4 int list (int A [], int n) // compare the size relationships of multiple d [j], and compare each time I is determined (d [j] + 1) comparison with 1 and initialization issues with d [I] 5 {int * d = (int *) malloc (sizeof (int) * n); // create a dynamic array, d is the array name 6 int j, I; 7 int len = 1; // d [n] indicates the longest non-descending subsequence before the nth Element 8 for (I = 0; I <n; ++ I) // I is used to traverse each element in the array 9 {10 d [I] = 1; // each I has a definite longest non-descending subsequence, therefore, you need to initialize d [I] 11 for (j = 0; j <I; ++ j) each time. // I is determined to be a value, in this case, we need to traverse the numbers before comparing I and I (j) 12 {13 if (A [j] <= A [I] & d [j] + 1> d [I]) // before I (j) the largest d [I] is required for multiple I-hours, similar to (a> max, max = a) 14 d [I] = d [j] + 1; 15} 16 if (d [I]> len) len = d [I]; // len returns the maximum value 17} 18 free (d); 19 20 return len; 21} 22 int main (void) 23 24 {25 26 int A [Max]; 27 int I; 28 for (I = 0; I <Max; I ++) 29 {30 scanf ("% d", & A [I]); 31} 32 printf ("the maximum length of the array's non-descending subsequence is % d \ n ", list (A, 10); 33 return 0; 34}