编程算法 - 背包问题(三种动态规划) 代码(C)

背包问题(三种动态规划) 代码(C)

本文地址: http://blog.csdn.net/caroline_wendy

题目参考: http://blog.csdn.net/caroline_wendy/article/details/37912949

可以用动态规划(Dynamic Programming, DP)求解, 可以通过记忆化搜索推导出递推式, 可以使用三种不同的方向进行求解.

动态规划主要是状态转移, 需要理解清晰.

代码:

 /*
  * 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]; //默认初始化为0
 public:
     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]<=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;
 }

输出:

 result = 7

节省空间, 可以使用1维数组的动态规划.

代码:

 /*
  * 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];
 public:
     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;
 }

输出:

 result = 7