目录

一、题目

1、题目描述

2、输入输出

2.1输入

2.2输出

3、原题链接

二、解题报告

1、思路分析

2、复杂度

3、代码详解

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一、题目

1、题目描述

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test, 1 day for 1 point. And as you know, doing homework always takes a long time. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

2、输入输出

2.1输入

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=15) which indicate the number of homework. Then N lines follow. Each line contains a string S(the subject's name, each string will at most has 100 characters) and two integers D(the deadline of the subject), C(how many days will it take Ignatius to finish this subject's homework).

Note: All the subject names are given in the alphabet increasing order. So you may process the problem much easier.

2.2输出

Output

For each test case, you should output the smallest total reduced score, then give out the order of the subjects, one subject in a line. If there are more than one orders, you should output the alphabet smallest one.

3、原题链接

Problem - 1074 (hdu.edu.cn)

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二、解题报告

1、思路分析

从数据量上会往状压dp上想

我们用二进制位表示任务是否完成，那么我们最终状态是确定的

如果有n个任务，那么我们的最终状态就是(1 << n) - 1，我们记为ed

对于ed而言，代表n个任务都已经完成，它可以由n个前驱状态转移而来

假如n = 3，那么ed = 111(2)，那么可以由011、101、110三个状态转移，分别代表最后完成的任务为任务1、2、3

那么对于011，101，110而言，同样可以由前驱状态转移

那么我们自顶向下进行状态转移即可

2、复杂度

时间复杂度：O(1<<N) 空间复杂度：O(1<<N)

3、代码详解

​

#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <stack>
using namespace std;
const int N = 20, inf = 0x3f3f3f3f;
struct state
{
    int pre, id, t, s;
} f[1 << N];
int cost[N], dead[N], n, tot;
string lessons[N];
void solve()
{
    cin >> n, tot = 1 << n, memset(f, 0, sizeof f);
    for (int i = 0; i < n; i++)
        cin >> lessons[i] >> dead[i] >> cost[i];
    for (int i = 1; i < tot; i++)
    {
        f[i].s = inf;
        for (int j = n - 1; j >= 0; j--)
        {
            if (i & (1 << j))
            {
                int last = i - (1 << j);
                int c = max(0, f[last].t + cost[j] - dead[j]);
                if (f[last].s + c < f[i].s)
                    f[i] = {last, j, f[last].t + cost[j], f[last].s + c};
            }
        }
    }
    cout << f[--tot].s << '\n';
    stack<int> s;
    while (f[tot].t)
    {
        s.emplace(f[tot].id), tot = f[tot].pre;
    }
    while (s.size())
        cout << lessons[s.top()] << '\n', s.pop();
}
int main()
{
    //freopen("in.txt", "r", stdin);
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int _ = 1;
    cin >> _;
    while (_--)
        solve();
    return 0;
}