题干：
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:
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
Hint
For the first case, Edward should select the first and second courses.
For the second case, Edward should select no courses.
解题报告：
容易观察到对于一个给定的ci，hi肯定是越大越好的，所以我们可以转化成容量是sum(ci)，价值分别是hi的01背包问题，这和那道蓝桥有异曲同工之妙。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int dp[MAX];
int h[MAX],c[MAX];
int main()
{int t,n;cin>>t;while(t--) {scanf("%d",&n);int m=0;for(int i = 1; i<=n; i++) {scanf("%d%d",h+i,c+i);m += c[i];}for(int i = 0; i<=m; i++) dp[i] = 0;for(int i = 1; i<=n; i++) {for(int j = m; j>=c[i]; j--) {dp[j] = max(dp[j],dp[j-c[i]]+h[i]);}}ll ans = 0;for(int i = 0; i<=m; i++) {ans = max(ans,1LL*dp[i]*dp[i]-1LL*dp[i]*i-1LL*i*i); }printf("%lld\n",ans);} return 0 ;
}
