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)

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

动态规划主要是状态转移, 须要理解清晰.

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, 能够使用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