计蒜客Dawn-K's water-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_44488944/article/details/103263901

本文介绍了一个基于完全背包问题的算法，用于解决如何在有限条件下购买矿物水以达到最少花费且重量不低于指定值的问题。通过使用动态规划，文章详细阐述了算法的实现过程，包括初始化、状态转移方程和求解步骤。

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Dawn-K recently discovered a very magical phenomenon in the supermarket of Northeastern University: The large package is not necessarily more expensive than the small package.

On this day, Dawn-K came to the supermarket to buy mineral water, he found that there are nn types of mineral water, and he already knew the price pp and the weight cc (kg) of each type of mineral water. Now Dawn-K wants to know the least money aa he needs to buy no less than mm kilograms of mineral water and the actual weight bb of mineral water he will get. Please help him to calculate them.

Input
The input consists of multiple test cases, each test case starts with a number n (1≤n≤10³) – the number of types, and mm (1≤m≤10⁴
) – the least kilograms of water he needs to buy. For each set of test cases, the sum of nn does not exceed 5e4.

Then followed n lines with each line two integers pp (1≤p≤10⁹) – the price of this type, and cc (1≤c≤10⁴) – the weight of water this type contains.

Output
For each test case, you should output one line contains the minimum cost aa and the weight of water Dawn-K will get bb. If this minimum cost corresponds different solution, output the maximum weight he can get.

(The answer aa is between 1 and 10⁹ , and the answer bb is between 1 and 10⁴
)

样例输入复制
3 3
2 1
3 1
1 1
3 5
2 3
1 2
3 3
样例输出复制
3 3
3 6

完全背包问题，一开始做的时候犯了好多低级错误。。。

#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=1010;
const int INF=0x3f3f3f3f;
long long p[N],c[N];
long long dp[10010];
int main()
{
long long n,m;
while(scanf("%lld %lld",&n,&m)!=EOF)
{
    memset(p,0,sizeof p);
    memset(c,0,sizeof c);
    memset(dp,INF,sizeof dp);//因为要求最小值，要将dp全部设为一个较大的值
    for(int i=1;i<=n;i++)
    {
        scanf("%lld %lld",&p[i],&c[i]);
    }
    dp[0]=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=c[i];j<=10000;j++)//dp[j]代表jkg的时候的价格
        {
            dp[j]=min(dp[j],dp[j-c[i]]+p[i]);//转移方程
        }
    }
    long long minw=INF;
    long long k=0;
   for(int i=m;i<=10000;i++)
   {
       if(minw>dp[i])//更新最小的价格，及价格最小时的质量
       {
            minw=dp[i];
            k=i;
       }
       else if(minw==dp[i]&&k<i)//如果最小价格相同时，买的越多越好
       {
           k=i;
       }
   }
   printf("%lld %lld\n",minw,k);
}
return 0;
}