hdoj2191 珍惜现在，感恩生活（01背包 || 多重背包）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=2191

思路

由于每种大米可能不止一袋，所以是多重背包问题，可以直接使用解决多重背包问题的方法，也可以将多重背包转化为01背包后求解。

代码

01背包：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 100 * 20 + 10;
int m[N], w[N];    //记录每袋大米的价格、重量
int dp[N];

int main()
{
    //freopen("hdoj2191.txt", "r", stdin);
    int t;
    cin >> t;
    while (t--)
    {
        int n, k;
        cin >> n >> k;
        int cnt = 0;    //共有cnt袋大米
        for (int i = 0; i < k; i++)
        {
            int p, h, c;
            cin >> p >> h >> c;
            for (int j = 0; j < c; j++)
            {
                m[cnt] = p;
                w[cnt] = h;
                cnt++;
            }
        }

        memset(dp, 0, sizeof(dp));
        for (int i = 0; i < cnt; i++)
        {
            for (int j = n; j >= m[i]; j--)
                dp[j] = max(dp[j], dp[j - m[i]] + w[i]);
        }
        cout << dp[n] << endl;
    }
    return 0;
}

多重背包：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 20 * 100 + 10;
int m[N], w[N], num[N];
int dp[N];

void zero_one_pack(int weight, int value, int capacity)
{
    for (int i = capacity; i >= weight; i--)    //逆序
        dp[i] = max(dp[i], dp[i - weight] + value);
}

void complete_pack(int weight, int value, int capacity)
{
    for (int i = weight; i <= capacity; i++)    //正序
        dp[i] = max(dp[i], dp[i - weight] + value);
}

void multiple_pack(int weight, int value, int amount, int capacity)
{
    if (weight*amount >= capacity)
        complete_pack(weight, value, capacity);
    else
    {
        int k = 1;
        while (k <= amount)
        {
            zero_one_pack(weight*k, value*k, capacity);
            amount -= k;
            k *= 2;
        }
        zero_one_pack(weight*amount, value*amount, capacity);
    }

}

int main()
{
    //freopen("hdoj2191.txt", "r", stdin);
    int t;
    cin >> t;
    while (t--)
    {
        int n, k;
        cin >> n >> k;
        for (int i = 0; i < k; i++)
            cin >> m[i] >> w[i] >> num[i];

        memset(dp, 0, sizeof(dp));
        for (int i = 0; i < n; i++)
            multiple_pack(m[i], w[i], num[i], n);
        cout << dp[n] << endl;
    }
    return 0;
}