Programming Algorithms-knapsack problem (three dynamic programming) code (C)

knapsack problem (three dynamic programming) code (C)

This address: Http://blog.csdn.net/caroline_wendy

Topic: http://blog.csdn.net/caroline_wendy/article/details/37912949

It can be solved by dynamic programming (dynamical programming, DP) , can be deduced by memory search , and can be solved in three different directions .

Dynamic planning is mainly the state transfer , need to understand clearly.

Code:

/* * main.cpp * * Created on:2014.7.17 * author:spike *//*eclipse CDT, gcc 4.8.1*/#include <stdio.h> #include <memory.h> #include <limits.h> #include <utility> #include <queue> #include <algorithm> Using namespace Std;class program {static const int max_n = 100;int n=4, w=5;int w[max_n] = {2,1,3,2}, v[max_n]={3,2,4,2}; int dp[max_n+1][max_n+1]; Default initialized to 0public:void solve () {for (int i=n-1; i>=0; i--) {for (int j=0; j<=w; ++j) {if (J<w[i]) {Dp[i][j] = dp[i+ 1][J];} else {Dp[i][j] = max (Dp[i+1][j], Dp[i+1][j-w[i]] + v[i]);}} printf ("result =%d\n", dp[0][w]);} void Solve1 () {for (int i=0, i<n; ++i) {for (int j=0; j<=w; ++j) {if (J<w[i]) {dp[i+1][j] = dp[i][j];} else {Dp[i +1][J] = max (Dp[i][j], dp[i][j-w[i]]+v[i]);}} printf ("result =%d\n", dp[n][w]);} void Solve2 () {for (Int. i=0; i<n; i++) {for (int j=0; j<=w; ++j) {dp[i+1][j] = max (Dp[i+1][j], dp[i][j]); if (j+w[i]& LT;=W) {Dp[i+1][j+w[i]] = max (Dp[i+1][j+w[i]], dp[i][j]+v[i]);}}} Printf ("result =%d\n", dp[n][w]);}}; int main (void) {program P;    P.solve2 (); return 0;}

Output:

result = 7

Save space and be able to use dynamic programming of 1-D arrays .

Code:

/* * main.cpp * *  Created on:2014.7.17 *      author:spike *//*eclipse CDT, gcc 4.8.1*/#include <stdio.h> #inclu De <memory.h> #include <limits.h> #include <utility> #include <queue> #include <algorithm >using namespace Std;class program {static const int max_n = 100;int n=4, w=5;int w[max_n] = {2,1,3,2}, v[max_n]={3,2,4 , 2};int dp[max_n+1];p ublic:void Solve () {memset (DP, 0, sizeof (DP)), for (Int. i=0; i<n; ++i) {for (int j=w; j>=w[i];- -J) {Dp[j] = max (Dp[j], dp[j-w[i]]+v[i]);}} printf ("result =%d\n", dp[w]);}; int main (void) {program P; P.solve ();    return 0;}

Output:

result = 7

Programming Algorithms-knapsack problem (three dynamic programming) code (C)