ZOJ 3956 Course Selection System

Course Selection System

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Time Limit: 1 Second      Memory Limit: 65536 KB

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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:

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

2310 15 12 1021 102 10

Sample Output

1910

Hint

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

For the second case, Edward should select no courses.

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Author: WANG, Yucheng
Source: The 17th Zhejiang University Programming Contest Sponsored by TuSimple

提示

题意：

从n个hi和ci中可不连续取任意个(意味着也可取0个)使得公式最后算出来的数值最大。

思路：

我们可以看做c固定不变的，要使得h尽可能的大，那么就可以转化为01背包问题。

示例程序

#include <stdio.h>#include <string.h>int main(){    int t,n;    long long i,i1,dp[50001],h,c,max;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        memset(dp,0,sizeof(dp));        for(i=1;n>=i;i++)        {            scanf("%lld %lld",&h,&c);            for(i1=50000;i1>=c;i1--)//这里的50000可优化为所有ci之和，减少循环次数，下同            {                if(dp[i1]<dp[i1-c]+h)                {                    dp[i1]=dp[i1-c]+h;                }            }        }        max=0;        for(i=0;50000>=i;i++)        {            if(max<dp[i]*dp[i]-dp[i]*i-i*i)            {                max=dp[i]*dp[i]-dp[i]*i-i*i;            }        }        printf("%lld\n",max);    }    return 0;}