zoj3956 Course Selection System (背包)

最新推荐文章于 2019-05-02 15:00:10 发布

原创最新推荐文章于 2019-05-02 15:00:10 发布 · 458 阅读

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本文介绍了一个课程选择系统的优化问题，学生需在有限的课程中选择最佳组合以最大化学期舒适度。通过将问题转化为背包问题，并利用动态规划求解最优解。

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Course Selection System

Time Limit: 1 Second Memory Limit: 65536 KB

There are n courses in the course selection system of Marjar University. The i-th course is described by two values: happiness H_i and credit C_i. If a student selects m courses x₁, x₂, ..., x_m, then his comfort level of the semester can be defined as follows:

$(\sum_{i=1}^{m} H_{x_i})^2-(\sum_{i=1}^{m} H_{x_i})\times(\sum_{i=1}^{m} C_{x_i})-(\sum_{i=1}^{m} C_{x_i})^2$

Edward, a student in Marjar University, wants to select some courses (also he can select no courses, then his comfort level is 0) to maximize his comfort level. Can you help him?

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains a integer n (1 ≤ n ≤ 500) -- the number of cources.

Each of the next n lines contains two integers H_i and C_i (1 ≤ H_i ≤ 10000, 1 ≤ C_i ≤ 100).

It is guaranteed that the sum of all n does not exceed 5000.

We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.

Output

For each case, you should output one integer denoting the maximum comfort.

Sample Input

Sample Output

191
0

Hint

For the first case, Edward should select the first and second courses.

For the second case, Edward should select no courses.

题解：

做的时候没有发现数据范围可以弄成一个背包，还想着贪心去做，结果赛后才发现500*100完全可以开得下，这样一来就可以当做一个背包问题来做了，复杂度是O（MN），转移方程是q[j-cost[i]]=max(q[j-cost[i]],q[j]-h[i]);

代码如下：

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

int q[50005];
int cost[505];
int h[505];

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        memset(q,-1,sizeof(q));
        int n;
        int sum1=0;
        int sum2=0;
        cin>>n;
        for(int i=0;i<n;i++)
        {
            scanf("%d %d",&h[i],&cost[i]);
            sum1+=h[i];
            sum2+=cost[i];
        }
        q[sum2]=sum1;
        long long ans=-1;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<=50000;j++)
            {
                if(q[j]==-1)
                    continue;
                q[j-cost[i]]=max(q[j-cost[i]],q[j]-h[i]);
            }
        }
        for(int i=0;i<50001;i++)
        {
            if(q[i]==-1)
                continue;
            ans=max(ans,(long long)q[i]*(long long)q[i]-(long long)i*(long long)q[i]-(long long)i*(long long)i);
        }
        cout<<ans<<endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}