全然背包问题 代码(C)

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

题目: 有n个重量和价值分别为w,v的物品, 从这些物品中挑选出总重量不超过W的物品, 求全部挑选方案中价值总和的最大值. 

*每件物品能够挑选随意多件.

动态规划: 每次选取最大的组合, 增加到数组, 第一种时间复杂度O(nW^2), 另外一种时间复杂度O(nW).

解法1, max()部分表明, 要么来源于上面, 要么来源于前面.

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];

public:

	void solve() {

		for (int i=0; i<n; ++i) {

			for (int j=0; j<=W; j++) {

				for (int k=0; k*w[i] <= j; k++) {

					dp[i+1][j] = max(dp[i+1][j], dp[i][j-k*w[i]]+k*v[i]);

				}

			}

		}

		printf("result = %d\n", dp[n][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+1][j-w[i]]+v[i]);

				}

			}

		}

		printf("result = %d\n", dp[n][W]);

	}

};

int main(void)

{

	Program P;

	P.solve1();

    return 0;

}

输出:

result = 10

为了节约内存, 能够使用一维数组进行求解.

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=3, W=7;

	int w[MAX_N] = {3,4,2}, v[MAX_N]={4,5,3};

	int dp[MAX_N+1];

public:

	void solve() {

		memset(dp, 0, sizeof(dp));

		for (int i=0; i<n; ++i) {

			for (int j=w[i]; j<=W; 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 = 10

能够讲两个数组滚动使用, 节省内存.

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=3, W=7;

	int w[MAX_N] = {3,4,2}, v[MAX_N]={4,5,3};

	int dp[2][MAX_N+1];

public:

	void solve() {

		memset(dp,0,sizeof(dp));

		for (int i=0; i<n; ++i) {

			for (int j=0; j<=W; j++) {

				if (j<w[i]) {

					dp[(i+1)&1][j] = dp[i&1][j];

				} else {

					dp[(i+1)&1][j] = max(dp[i&1][j], dp[(i+1)&1][j-w[i]]+v[i]);

				}

			}

		}

		printf("result = %d\n", dp[n&1][W]);

	}

};

int main(void)

{

	Program P;

	P.solve();

    return 0;

}

输出:

result = 10

当重量可选范围非常大, 价值可选范围小的时候, 能够使依据价值进行动态规划.

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;

	static const int MAX_V = 4;

	const int INF = INT_MAX>>2;

	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*MAX_V+1];

public:

	void solve() {

		fill(dp[0], dp[0]+MAX_N*MAX_V+1, INF);

		dp[0][0] = 0;

		for (int i=0; i<n; ++i) {

			for (int j=0; j<=MAX_N*MAX_V; j++) {

				if (j<v[i]) {

					dp[i+1][j] = dp[i][j];

				} else {

					dp[i+1][j] = min(dp[i][j], dp[i][j-v[i]]+w[i]);

				}

			}

		}

		int res = 0;

		for (int i=0; i<=MAX_N*MAX_V; ++i) if (dp[n][i]<=W) res = i;

		printf("result = %d\n", res);

	}

};

int main(void)

{

	Program P;

	P.solve();

    return 0;

}

输出:

result = 7