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;
}

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;
}

result = 7