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hdoj 5501 The Highest Mark 【贪心 + 0-1背包】

2015-10-11 17:07 387 查看

The Highest Mark

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 324    Accepted Submission(s): 129


Problem Description

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

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:

First to solve problem 2 and
then solve problem 1 he
will get 88 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

 

题意：给出n道题和每道题的分值、每分钟减少的分值、解决该问题需要的时间，问你在时限m分钟内，可以得到的最大分值。假设你可以解决所有问题。

思路：以为就是0-1背包，没想到测试数据都不过，加个贪心预处理一下就ok了。对于每道题，可选择做或不做，得到状态转移方程dp[j] = max(dp[j], dp[j-num[i].C] + num[i].A - j*num[i].B) (0 <= j <= m)。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 3000+10
#define INF 0x3f3f3f3f
using namespace std;
int dp[MAXN];
struct rec{
int A, B, C;
};
rec num[1010];
bool cmp(rec a, rec b){
return a.B * b.C > a.C * b.B;
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
int n, m;
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++)
scanf("%d%d%d", &num[i].A, &num[i].B, &num[i].C);
sort(num, num+n, cmp);
memset(dp, 0, sizeof(dp));
for(int i = 0; i < n; i++)
for(int j = m; j >= num[i].C; j--)
dp[j] = max(dp[j], dp[j-num[i].C] + num[i].A - j * num[i].B);
int ans = 0;
for(int i = 0; i <= m; i++)
ans = max(dp[i], ans);
printf("%d\n", ans);
}
return 0;
}

                                            

                                        

                        
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