hdoj 5501 The Highest Mark 【贪心 + 0-1背包】_sample input 1 4 10 30 2 1 50 3 2 80 4 8 110 5 9-优快云博客

本文讨论了一个比赛中的得分优化问题，参与者需要在限定时间内解决不同难度的问题，并通过贪心算法和0-1背包策略来实现最高分数的获取。

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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

Sample Output

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;
}