zoj 3956

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:

(∑i=1mHxi)2−(∑i=1mHxi)×(∑i=1mCxi)−(∑i=1mCxi)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
Hint

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

For the second case, Edward should select no courses.

枚举：学分从1到学分和的最大H值,枚举时可能枚举到没有的分数，写出数据会发现不影响，因为大的学分是值小如
1
2
100 1
10 5

dp 1 2 3 4 5 6 H值 100 100 100 100 100 110

没有学分2，3，4，但H对应的C为1最大
H值用01背包
把学分和当背包体积
H是价值
C是物品体积

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>using namespace std;int main(int argc, char *argv[]){    int T,n,w[509],c[509],v;    long long maxA,dp[50009];    scanf("%d",&T);    while (T--) {        v = 0, maxA = 0;        memset(dp, 0, sizeof(dp));        scanf("%d",&n);        for (int i = 1; i <= n; i++) {            scanf("%d%d",&w[i],&c[i]);            v += c[i];         }        for (int i = 1; i <= n; i++) {            for (int j = v; j >= c[i]; j--) {                dp[j] = max(dp[j], dp[j-c[i]] + w[i]);            }        }        for (int i = 1; i <= v; i++) {            maxA = max(maxA, dp[i]*dp[i] - dp[i]*i - i*i);        }        printf("%lld\n",maxA);    }    return 0;}