hdoj2191 珍惜现在，感恩生活

Posted 2020-10-17 ColdCode

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：

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 using namespace std;
 6 
 7 const int N = 100 * 20 + 10;
 8 int m[N], w[N];    //记录每袋大米的价格、重量
 9 int dp[N];
10 
11 int main()
12 {
13     //freopen("hdoj2191.txt", "r", stdin);
14     int t;
15     cin >> t;
16     while (t--)
17     {
18         int n, k;
19         cin >> n >> k;
20         int cnt = 0;    //共有cnt袋大米
21         for (int i = 0; i < k; i++)
22         {
23             int p, h, c;
24             cin >> p >> h >> c;
25             for (int j = 0; j < c; j++)
26             {
27                 m[cnt] = p;
28                 w[cnt] = h;
29                 cnt++;
30             }
31         }
32 
33         memset(dp, 0, sizeof(dp));
34         for (int i = 0; i < cnt; i++)
35         {
36             for (int j = n; j >= m[i]; j--)
37                 dp[j] = max(dp[j], dp[j - m[i]] + w[i]);
38         }
39         cout << dp[n] << endl;
40     }
41     return 0;
42 }

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;
}