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动态规划的用法——01背包问题

2015-01-10 00:00 801 查看
动态规划的用法——01背包问题

问题主题：著名的01背包问题
问题描述： 有n个重量和价值分别为wi、vi的物品，现在要从这些物品中选出总重量不超过W的物品，求所有挑选方案中的价值最大值。 限制条件： 1<=N<=100 1<=wi 、vi<=100 1<=wi<=10000
样例： 输入 N=4 W
= {2, 1, 3, 2} V
= {3, 2, 4, 2} 输出 W = 5(选择0，1，3号)

【解法一】

解题分析：

用普通的递归方法，对每个物品是否放入背包进行搜索

程序实现:

C++
#include <stdio.h>
#include <tchar.h>
#include <queue>
#include "iostream"

using namespace std;

const int N = 4;
const int W = 5;
int weight
= {2, 1, 3, 2};
int value
= {3, 2, 4, 2};
int solve(int i, int residue)
{
int result = 0;
if(i >= N)
return result;
if(weight[i] > residue)
result = solve(i+1, residue);
else
{
result = max(solve(i+1, residue), solve(i+1, residue-weight[i]) + value[i]);
}

}

int main() {
int result = solve(0, W);
cout << result << endl;
return 0;
}


【解法二】

解题分析：

用记录结果再利用的动态规划的方法，上面用递归的方法有很多重复的计算，效率不高。我们可以记录每一次的计算结果，下次要用时再直接去取，以提高效率

程序实现:

C++
#include <stdio.h>
#include <tchar.h>
#include <queue>
#include "iostream"

using namespace std;

const int N = 4;
const int W = 5;
int weight
= {2, 1, 3, 2};
int value
= {3, 2, 4, 2};
int record
[W];
void init()
{
for(int i = 0; i < N; i ++)
{
for(int j = 0; j < W; j ++)
{
record[i][j] = -1;
}
}
}

int solve(int i, int residue)
{
if(-1 != record[i][residue])
return record[i][residue];
int result = 0;
if(i >= N)
return result;
if(weight[i] > residue)
{
record[i + 1][residue] = solve(i+1, residue);

}
else
{
result = max(solve(i+1, residue), solve(i+1, residue-weight[i]) + value[i]);
}
return record[i + 1][residue] = result;
}

int main() {
init();
int result = solve(0, W);
cout << result << endl;
return 0;
}


Java
package greed;

/**
* User: luoweifu
* Date: 14-1-21
* Time: 下午5:13
*/
public class Knapsack {
private int maxWeight;
private int[][] record;
private Stuff[] stuffs;

public Knapsack(Stuff[] stuffs, int maxWeight) {
this.stuffs = stuffs;
this.maxWeight = maxWeight;
int n = stuffs.length + 1;
int m = maxWeight+1;
record = new int
[m];
for(int i = 0; i < n; i ++) {
for(int j = 0; j < m; j ++) {
record[i][j] = -1;
}
}
}
public int solve(int i, int residue) {
if(record[i][residue] > 0) {
return record[i][residue];
}
int result;
if(i >= stuffs.length) {
return 0;
}
if(stuffs[i].getWeight() > residue) {
result = solve(i + 1, residue);
} else {
result = Math.max(solve(i + 1, residue),
solve(i + 1, residue - stuffs[i].getWeight()) + stuffs[i].getValue());
}
record[i][residue] = result;
return result;
}

public static void main(String args[]) {
Stuff stuffs[] = {
new Stuff(2, 3),
new Stuff(1, 2),
new Stuff(3, 4),
new Stuff(2, 2)
};
Knapsack knapsack = new Knapsack(stuffs, 5);
int result = knapsack.solve(0, 5);
System.out.println(result);
}
}

class Stuff{
private int weight;
private int value;

public Stuff(int weight, int value) {
this.weight = weight;
this.value = value;
}

int getWeight() {
return weight;
}

void setWeight(int weight) {
this.weight = weight;
}

int getValue() {
return value;
}

void setValue(int value) {
this.value = value;
}
}


 
 算法复杂度：
时间复杂度O(NW)
                                            

                                        

                        
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                         标签： 
                                                程序设计
                                                算法
                                                动态规划
                                                

                        

                    


                

            


            

                
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