状压dp,HDU1074.Doing Homework

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


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)


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

复制代码
cpp 复制代码
#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;
}
相关推荐
BB_CC_DD18 分钟前
四. 以Annoy算法建树的方式聚类清洗图像数据集,一次建树,无限次聚类搜索,提升聚类搜索效率。(附完整代码)
深度学习·算法·聚类
梁下轻语的秋缘2 小时前
每日c/c++题 备战蓝桥杯 ([洛谷 P1226] 快速幂求模题解)
c++·算法·蓝桥杯
CODE_RabbitV2 小时前
【深度强化学习 DRL 快速实践】逆向强化学习算法 (IRL)
算法
mit6.8242 小时前
[贪心_7] 最优除法 | 跳跃游戏 II | 加油站
数据结构·算法·leetcode
keep intensify2 小时前
通讯录完善版本(详细讲解+源码)
c语言·开发语言·数据结构·算法
shix .2 小时前
2025年PTA天梯赛正式赛 | 算法竞赛,题目详解
数据结构·算法
风铃儿~3 小时前
Java面试高频问题(26-28)
java·算法·面试
wuqingshun3141593 小时前
蓝桥杯 4. 卡片换位
算法·职场和发展·蓝桥杯
江沉晚呤时3 小时前
深入了解C# List集合及两种常见排序算法:插入排序与堆排序
windows·sql·算法·oracle·c#·排序算法·mybatis
Eric.Lee20213 小时前
数据集-目标检测系列- F35 战斗机 检测数据集 F35 plane >> DataBall
人工智能·算法·yolo·目标检测·计算机视觉