題目連結

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 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.

【題意】

Edward從n門課中選m門，沒門課有相應的績點c，並且Edward可以從中獲取相應的快樂值h，輸出一個使題中等式最大的答案。

【解題思路】

令a=sigma(hi)，b=sigma(ci)，那麼上式可以化簡為f=a^2-ab-b^2=a(a-b)-b^2。

那麼當b一定時，a越大f越大，那麼就變成01揹包問題啦。

【程式碼】

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL h[505],c[505],f[500*100+5];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,C=0;
        memset(f,0,sizeof(f));
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&h[i],&c[i]);
            C+=c[i];
        }
        for(int i=0;i<n;i++)
            for(int j=C;j>=c[i];j--)
                f[j]=max(f[j],f[j-c[i]]+h[i]);
        LL ans=0;
        for(int i=0;i<=C;i++)
            ans=max(ans,f[i]*f[i]-f[i]*i-i*i);
        printf("%lld\n",ans);
    }
    return 0;
}