Course Selection System 【01背包】

There are n courses in the course selection system of Marjar University. The i-th course is described by two values: happiness Hi and credit Ci. If a student selects m courses x1, x2, …, xm, 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 Hi and Ci (1 ≤ Hi ≤ 10000, 1 ≤ Ci ≤ 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
2
3
10 1
5 1
2 10
2
1 10
2 10
Sample Output
191
0
题意很好理解，课程选或者不选，可以向01背包上考虑，但是没有找到固定的东西
这道题目选择C固定，因为C比较小，而且，可以知道当C固定的时候，要尽可能的选择较大的h。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
struct node
{
    long long int c, h;
}t[12311];
long long int Max(long long int a, long long int b)
{
    if(a>b)
        return a;
    return b;
}
long long int dp[560007];//c*n的数量级
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        long long int sum = 0;
        for(int i=0;i<n;i++)
        {
            scanf("%lld %lld", &t[i].h, &t[i].c);
            sum += t[i].c;
        }
        memset(dp, 0, sizeof(dp));
        for(int i=0;i<n;i++)
        {
            for(int j=sum;j>=t[i].c;j--)//c固定，使h尽可能大
            {
                dp[j] = Max(dp[j-t[i].c]+t[i].h, dp[j]);
            }
        }
        long long int maxn = 0;
       for(int i=0;i<=sum;i++)
       {
           maxn = Max(maxn, dp[i]*dp[i]-dp[i]*i-i*i);//题目所给公式
       }
       printf("%lld\n", maxn);
    }
    return 0;
}