source

Description

The SDOI in 20452045 is far from what it was been 3030 years ago. Each competition has tt minutes and nn problems.

The ithith problem with the original mark of Ai(Ai≤106)Ai(Ai≤106),and it decreases BiBi by each minute. It is guaranteed that it does not go to minus when the competition ends. For example someone solves the ithith problem after xx minutes of the competition beginning. He/She will get Ai−Bi∗xAi−Bi∗x marks.

If someone solves a problem on xx minute. He/She will begin to solve the next problem on x+1x+1 minute.

dxy who attend this competition with excellent strength, can measure the time of solving each problem exactly.He will spend Ci(Ci≤t)Ci(Ci≤t) minutes to solve the ith problem. It is because he is so godlike that he can solve every problem of this competition. But to the limitation of time, it's probable he cannot solve every problem in this competition. He wanted to arrange the order of solving problems to get the highest mark in this competition.
Input
There is an positive integer T(T≤10)T(T≤10) in the first line for the number of testcases.(the number of testcases with n>200n>200 is no more than 55)

For each testcase, there are two integers in the first line n(1≤n≤1000)n(1≤n≤1000) and t(1≤t≤3000)t(1≤t≤3000) for the number of problems and the time limitation of this competition.

There are nn lines followed and three positive integers each line Ai,Bi,CiAi,Bi,Ci. For the original mark,the mark decreasing per minute and the time dxy of solving this problem will spend.

Hint:
First to solve problem 22 and then solve problem 11 he will get 8888 marks. Higher than any other order.
Output
For each testcase output a line for an integer, for the highest mark dxy will get in this competition.

Sample Input

1
4 10
110 5 9
30 2 1
80 4 8
50 3 2

Sample Output

88

结题报告

#include<cstdio>
#include<algorithm>
#include<string.h>
using namespace std;
struct Node{
    int A;
    int B;
    int C;
    bool operator <(const Node &t)
    {
        return (long long)B*(t.C)>(long long)(t.B)*C;

    }
} arr[3500];
int dp[3500];
int main() {

    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,t;
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&t);
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d",&arr[i].A,&arr[i].B,&arr[i].C );

        }
        sort(arr+1,arr+n+1);
        for(int i=1;i<=n;i++){
            for(int j=t;j>=arr[i].C;j--)
            {
                dp[j]=max(dp[j],dp[j-arr[i].C]+arr[i].A-arr[i].B*j);
            }
        }
         int ans = 0;
        for(int i = 0; i <= t; i++) {
            ans = max(ans, dp[i]);
        }
        printf("%d\n", ans);
    }
}