0-1背包问题的动态规划解法之二位数组

#include <iostream>
using namespace std;

int num[4] = { 2, 2, 4, 6 };
int f()
{
    bool state[5][6];  //在linux中注意用memset(state,false,sizeof(state))初始化；这里可以不用，因为自动被初始化为false
    state[0][0] = true;

    for (int i = 1; i < 5; i++)   
    {
        for (int j = 0; j <=5; j++)    //物品不装入背包
        {
            if (state[i - 1][j] == true)
                state[i][j] = true;
        }

        if (num[i] > 5)  //如果本物品重量超过可容纳重量，则继续下一个循环
            continue;

        for (int j = 0; j <= 5 - num[i]; j++) //把物品装入背包
        {

            if (state[i - 1][j] == true)
                state[i][j + num[i]] = true;
        }
    }

    for (int j = 5; j >= 0; j--)  //输出
    {
        if (state[4][j] == true)
            return j;
    }    
    return 0;
 }

 int main()
 {
     int i = f();
     cout << i << endl;     
     system("pause");
     return i;
 }