PAT 1002. Business (35)

As the manager of your company, you have to carefully consider, for each project, the time taken to finish it, the deadline, and the profit you can gain, in order to decide if your group should take this project. For example, given 3 projects as the following:

Project[1] takes 3 days, it must be finished in 3 days in order to gain 6 units of profit.
Project[2] takes 2 days, it must be finished in 2 days in order to gain 3 units of profit.
Project[3] takes 1 day only, it must be finished in 3 days in order to gain 4 units of profit.
You may take Project[1] to gain 6 units of profit. But if you take Project[2] first, then you will have 1 day left to complete Project[3] just in time, and hence gain 7 units of profit in total. Notice that once you decide to work on a project, you have to do it from beginning to the end without any interruption.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N(<=50), and then followed by N lines of projects, each contains three numbers P, L, and D where P is the profit, L the lasting days of the project, and D the deadline. It is guaranteed that L is never more than D, and all the numbers are non-negative integers.

Output Specification:

For each test case, output in a line the maximum profit you can gain.

Sample Input:
4
7 1 3
10 2 3
6 1 2
5 1 1
Sample Output:
18

看题意可以知道要么是暴力搜索，要么是dp
但是N是50，很显然dfs是会超时的，一般来说n在10左右dfs才不会超时。
那就肯定是DP嘛，不可能还有别的做法吧。
要是贪心题的话N必然是1e5左右的大小。
其实数据规模能够暗示很多东西。

那么既然是dp，就肯定是背包DP。
但是题目有个限定，不能超过DDL。

然后我头很铁，就当成普通01背包来做，做的时候判是否超过DDL，如果超过的话，就continue
然后开心的发现样例都错了
然后抱着尝试的态度按照DDLsort了一下，因为对于同样value，同样重量的物品，肯定是ddl越前面应该先去考虑嘛
然后交了一发，WA了

然后参考了别人的博客，发现不能简单的输出dp终点。
得在前面定义一个变量，时刻检测背包的值，并且输出最大值。
举个简单的例子
3
10 1 200
10 1 99
10 1 99
背包大小肯定是选择所有DDL中最大的一个作为背包容量，就是200，但是dp[200]是10，但是答案是30
原因在于你在当前背包容量超过DDL的时候，是continue的
好吧其实感觉自己也讲不太明白，毕竟DP题就是靠感觉的。。。。

#include <bits/stdc++.h>
using namespace std;
struct node{
	int value;
	int last;
	int ddl;
	bool operator< (const node& b) const
	{
		if (ddl != b.ddl) return ddl < b.ddl;//靠猜的，不会证 
 	   else return last < b.last;
	}
}arr[100];
int dp[100000];
int main(void)
{
	int n;
	scanf("%d", &n);
	int i, j;
	int M = -1;
	for (i = 1; i <= n; i++)
	{
		scanf("%d%d%d", &arr[i].value, &arr[i].last, &arr[i].ddl);
		M = max(M, arr[i].ddl);//选取最大DDL作为背包，因为
							   //你超过这个时间是没有意义的
							   //不会有任何rpoject可以安插在这个更大的背包里 
	}
	sort(arr+1, arr+1+n);
	int res = -1;//定义变量 
	for (i = 1; i <= n; i++)//01背包模型 
	{
		for (j = M; j >= arr[i].last; j--)
		{
			if (j > arr[i].ddl) continue;
			dp[j] = max(dp[j], dp[j-arr[i].last] + arr[i].value);
			res = max(res, dp[j]);//时刻更新res 
		}
	}
	cout << res;
}