The Preliminary Contest for ICPC Asia Shenyang 2019 C。Dawn-K's water

最新推荐文章于 2020-01-16 01:59:05 发布

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最新推荐文章于 2020-01-16 01:59:05 发布

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分类专栏： dp

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本文探讨了一个有趣的现象：在购买商品时，大包装不一定比小包装更贵。通过具体案例分析，介绍了一种算法来帮助消费者找到购买指定重量商品时的最低成本及实际获取的重量，特别适用于大量购买矿泉水的情况。

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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 nn (1 \le n \le 10^31≤n≤103) -- the number of types, and mm (1 \le m \le 10^41≤m≤104) -- the least kilograms of water he needs to buy. For each set of test cases, the sum of nn does not exceed 5e45e4.

Then followed n lines with each line two integers pp (1 \le p \le 10^91≤p≤109) -- the price of this type, and cc (1 \le c \le 10^41≤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 11 and 10^9109, and the answer bb is between 11 and 10^4104)

样例输入复制

样例输出复制

3 3
3 6

完全背包的问题，一开始没有看到最重的水的上限1e4，然后两次完全背包就一直超时然后贪心最大性价比也一直错。后来队友A了才知道有个最大限制就把背包的最大容量看成1e4然后从m到1e4枚举钱数就行了。

编辑代码

#include <bits/stdc++.h>
using namespace std;
int v[50000],w[50000];
int dp[10009];
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        memset(dp,0x3f3f3f3f,sizeof(dp));
        dp[0]=0;
        for(int i=1;i<=n;i++)
            cin>>v[i]>>w[i];
        for(int i=1;i<=n;i++)
        {
            for(int j=w[i];j<=1e4;j++)
                dp[j]=min(dp[j],dp[j-w[i]]+v[i]);
        }
        int ans1=0x3f3f3f3f,ans2=m;
        for(int i=m;i<=1e4;i++)
        {
            if(dp[i]<=ans1)
            {
                ans1=dp[i],ans2=i;
            }
        }
        cout<<ans1<<' '<<ans2<<endl;
    }
}