The Highest Mark_hdu5501_dp+贪心

Problem Description

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The SDOI in 2045 is far from what it was been 30 years ago. Each competition has t minutes and n problems.

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

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

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There is an positive integer T(T≤10) in the first line for the number of testcases.(the number of testcases with n>200 is no more than 5)

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

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

Hint:

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First to solve problem 2 and then solve problem 1 he will get 88 marks. Higher than any other order.

Output

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For each testcase output a line for an integer, for the highest mark dxy will get in this competition.

Source

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BestCoder Round #59 (div.1)

Analysis

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不难想出最优解肯定是连续做完某些题目，那么问题就是如何找到最优的做题顺序
考虑物品T1和T2
当T1先做，损失的价值为C1∗B1+B2∗(C1+C2)
当T2先做，损失的价值为C2∗B2+B1∗(C1+C2)
先贪心，拆开来去重按照B2∗C1与B1∗C2升序排序，就变成01背包了

Code

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#include <cstdio>#include <cstring>#include <algorithm>#define ll long long#define N 1501#define M 4001using namespace std;struct num{ll a,b,c;}t[N];ll f[M];ll max(ll x,ll y){return x<y?y:x;}int cmp(num x,num y){return x.c*y.b<x.b*y.c;}int main(){    int T;    scanf("%d",&T);    while (T--)    {        int n,m;        ll ans=0;        scanf("%d%d",&n,&m);        for (int i=1;i<=n;i++)            scanf("%lld%lld%lld",&t[i].a,&t[i].b,&t[i].c);        for (int i=0;i<=m+1;i++)            f[i]=0;        sort(t+1,t+n+1,cmp);        for (int i=1;i<=n;i++)            for (int j=m;j>=t[i].c;j--)            {                f[j]=max(f[j],f[j-t[i].c]+t[i].a-t[i].b*j);                ans=max(f[j],ans);            }        printf("%lld\n",ans);    }    return 0;}