hdu1114 Piggy-Bank

最新推荐文章于 2022-01-22 21:04:35 发布

原创最新推荐文章于 2022-01-22 21:04:35 发布 · 503 阅读

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背包问题

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本文介绍了一个使用完全背包算法解决的问题实例。通过初始化数组并利用完全背包的特性进行迭代更新，最终求解出背包能容纳的最大价值。代码中详细展示了如何处理物品的加入，并通过具体的例子演示了整个求解过程。

解题思路：完全背包

注意要初始 d[0] = 0

AC代码：

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

#define clr(p,v) memset(p,v,sizeof(p))
const int maxn = 10010 ;

int n, m, C;
int d[maxn];

void zeroOnePack(int cost, int weight)
{
    for (int i=m; i>=cost; --i)
        d[i] = max(d[i], d[i-cost]+weight);
}

void completePack(int cost, int weight)
{
    for (int i=cost; i<=m; ++i)
        d[i] = max(d[i], d[i-cost]+weight);
}

void multiPack(int cost, int weight, int amount)
{
    if (cost*amount >= m)
        completePack(cost, weight);
    else
    {
        int num = 1;
        while (num <= amount)
        {
            zeroOnePack(num*cost, num*weight);
            amount -= num;
            num <<= 1;
        }
        zeroOnePack(amount*cost, amount*weight);
    }
}

int main()
{
    scanf("%d", &C);
    while (C--)
    {
        //Input && Calculate
        int tt;
        scanf("%d%d%d", &tt, &m, &n);
        m -= tt;
        //init d[]
        for (int i=1; i<=m; ++i)
            d[i] = -500000010;
        d[0] = 0;
        
        for (int i=1; i<=n; ++i)
        {
            int p, w;
            scanf("%d%d", &p, &w);
            completePack(w, -p);
        }
        //output
        if (-d[m] != 500000010 && d[m])
            printf("The minimum amount of money in the piggy-bank is %d.\n", -d[m]);
        else
            puts("This is impossible.");
    }
    return 0;
}