Problem solving ideas of dynamic planning:
There are four steps:
1 create a new one-dimensional array or two-dimensional array, and determine the type of array according to the variables given in the question.
2 if it is a one-dimensional array, determine the value of the first element of the array.
If it is a two-dimensional array, determine the values of the first row and first column of the array.
3 find out the state transition equation.
4 return results.
Example 1: find the maximum discontinuous subsequence
Let L = < A1, A2,..., an > be a sequence of n different real numbers, and the increasing subsequence of L is such a subsequence Lin = < AK1, ak2,..., AKM >, where K1 < K2 <... < km and AK1 < ak2 <... < AKM. Find the maximum value of m.
Idea: first, set an array temp[i] to store the longest subsequence of I. Secondly, initialize temp[i]=, and then find the state transition equation: if for (J = 1; j<i; J + +) when arr [i] > arr [J], temp[i]=max(temp[i],temp[j]+1) when arr [i] < arr [J], then temp[i] is not updated
#include <iostream> #include <bits/stdc++.h> #include <string> using namespace std; //test data /* 4 2 1 3 50 */ // test result: 3 //Longest discontinuous subsequence problem int main() { int n; int arr[101]; //Define array storage sequence cin>>n; for(int i=1;i<=n;i++) { cin>>arr[i]; } int temp[101]; //The first step of dynamic programming is to establish a one-dimensional array for(int i=1;i<=n;i++) //Initialize element temp[i]=1; for(int i=2;i<=n;i++) //state transition { for(int j=1;j<i;j++) { if(arr[i]>arr[j]) { temp[i]=max(temp[i],temp[j]+1); } } } //Find the maximum value of temp array int maxt=0; for(int i=1;i<=n;i++) { if(maxt<temp[i]) { maxt=temp[i]; } } cout<<maxt<<endl; return 0; }
Example 2: find the sum of continuous maximum subsequences
Given an integer array arr [], find a continuous subarray with the largest sum (the subarray contains at least one element), and return its maximum sum
Idea: first set an array temp[i] to store and; temp[1] is initialized to arr[1]; When temp[i-1] + arr [i] > arr [i], temp = temp[i-1] + arr [i], otherwise temp[i]=arr[i];
code:
#include <iostream> #include <bits/stdc++.h> #include <string> using namespace std; /*test data Input: 9 -2 1 -3 4 -1 2 1 -5 4 Output: 6 Explanation: continuous subsequence and maximum, 6 */ //Maximum subsequence sum int main() { int n; int arr[101]; //Store original array int temp[101]; //Storage and cin>>n; for(int i=1;i<=n;i++) { cin>>arr[i]; } temp[1]=arr[1]; //initialization for(int i=2;i<=n;i++) //state transition { if(temp[i-1]+arr[i]>arr[i]) temp[i]=temp[i-1]+arr[i]; else temp[i]=arr[i]; } //Find the maximum value of temp [] array int maxand=temp[1]; for(int i=2;i<=n;i++) { if(temp[i]>maxand) maxand=temp[i]; } cout<<maxand<<endl; return 0; }
Example 3: maximum common subsequence of two strings
Given two strings, the Longest Common Sequence of the two strings is solved. For example, string 1: BDCABA; String 2: ABCBDAB
Then the length of the longest common subsequence of the two strings is 4, and the longest common subsequence is BCBA
Idea: set a two-dimensional array dp[i][j] to represent the longest subsequence of coordinates (i,j). Initialize the first row and column. If x[i]=y[j], dp[i][j]=dp[i-1][y-1]+1;
Otherwise, dp[i][j]=max(dp[i-1][j], dp[i][j-1]);
code:
#include <iostream> #include <bits/stdc++.h> #include <string> using namespace std; //Maximum common subsequence int main() { string x; string y; cin>>x; cin>>y; int dp[21][21]; int len1=x.length(); int len2=y.length(); memset(dp,0,sizeof(dp)); for(int i=1;i<=len1;i++) { for(int j=1;j<=len2;j++) { if(x[i-1]==y[j-1]) { dp[i][j]=dp[i-1][j-1]+1; } else { dp[i][j]=max(dp[i-1][j],dp[i][j-1]); } } } cout<<dp[len1][len2]<<endl; int k=0; char temp[101]; int p=len1-1; int q=len2-1; while(p>=0&&q>=0) { if(dp[p+1][q+1]==dp[p][q]+1&&x[p]==y[q]) { temp[k++]=x[p]; p--; q--; } else if(dp[p+1][q+1]==dp[p][q+1]) { p--; } else { q--; } } for(int i=k-1;i>=0;i--) { cout<<temp[i]<<" "<<endl; } return 0; }
Example 4: change problem
Problem Description:
There are a pile of coins with denominations of 1,2,5,11,20,50. How many coins do you need to find the change with a total value of N units
**Idea: * * create a two-dimensional array, the ordinate represents the face value of coins, and the abscissa represents the change of j units. Initialize the first row and the first column. For each coin, if this coin is used, dp[i][j]=dp[i][j-p[i]]+dp[j], otherwise dp[i][j]=dp[i-1][j];
code:
#include <iostream> #include <bits/stdc++.h> #include <string> using namespace std; /*Test sample 3 1 3 4 4 */ // Output: 3 //Change problem /* In the first line, enter the number of face values n Enter face value arr [] On the third line, enter the amount of change to be found */ int main() { int n; int p; int arr[101]; int dp[21][21]; cin>>n; for(int i=1; i<=n; i++) { cin>>arr[i]; } cin>>p; memset(dp,0,sizeof(dp)); for(int i=1; i<=n; i++) { dp[i][0]=1; } for(int i=0; i<=p; i++) { if(i%arr[1]==0) dp[1][i]=1; else dp[1][i]=0; } for(int i=2; i<=n; i++) { for(int j=1; j<=p; j++) { if(j>=arr[i]) { dp[i][j]=dp[i-1][j]+dp[i][j-arr[i]]; } else { dp[i][j]=dp[i-1][j]; } } } cout<<dp[n][p]<<endl; return 0; }