首页 > 代码库 > hdu 1074 Doing Homework

hdu 1074 Doing Homework

2024-07-02 09:33:59   225人阅读

Problem Description
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.
Input
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.
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.
Sample Input
2
3
Computer 3 3
English 20 1
Math 3 2
3
Computer 3 3
English 6 3
Math 6 3
Sample Output
2
Computer
Math
English
3
Computer
English
Math
 

看到这么一个题，每个人都有一个念头，就是dfs，我也是，于是就写了一个dfs

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f

using namespace std;

const double pi=acos(-1.0);
const double eps=1e-8;
typedef pair<int,int>pii;

struct node
{
    char name[101];
    int deadline;
    int needtime;
}homework[15];

bool vis[16];

int path[16],p[16];
int reduce,n;

void dfs(int num,int nowreduce,int nowtime)
{
    if (num==n)
    {
        if (reduce>nowreduce)
        {
            reduce=nowreduce;
            for (int i=1;i<=n;i++) p[i]=path[i];
        }
        return;
    }
    if (nowreduce>=reduce) return;
    for (int i=1;i<=n;i++)
    if (!vis[i])
    {
        vis[i]=true;
        path[num+1]=i;
        int temp=nowreduce+nowtime+homework[i].needtime-homework[i].deadline;
        if (nowtime+homework[i].needtime<homework[i].deadline)temp=nowreduce;
        dfs(num+1,temp,nowtime+homework[i].needtime);
        vis[i]=false;
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    int t;
    scanf("%d",  &t);
    while (t--)
    {
        scanf("%d", &n);
        for (int i=1;i<=n;i++)
        {
            scanf("%s %d %d",&homework[i].name, &homework[i].deadline, &homework[i].needtime);
        }
        reduce=inf;
        memset(vis,0,sizeof(vis));
        dfs(0,0,0);
        printf("%d\n",reduce);
        for (int i=1;i<=n;i++) printf("%s\n",homework[p[i]].name);
    }

    //fclose(stdin);
    return 0;
}

还很高兴自己写了那么一个好的dfs ，谁知一提交，tle了，正常，回头算一下，最坏的情况下为15！，肯定超时，再看n==15；知道肯定是集合上的dp，但是不知道怎么保存路径，和怎么转移方程，于是看了看别人的代码，模仿着写了这个，正所谓一次不会我学，两次不会我还学，我就不相信我会永远不会

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<stack>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f

using namespace std;

const double pi=acos(-1.0);
const double eps=1e-8;
typedef pair<int,int>pii;

struct Home
{
    char name[101];
    int deadline,needtime;
}homework[16];

struct node
{
    int pre,score,time,last;
}dp[1<<16];


int main()
{
    //freopen("in.txt","r",stdin);

    int t,reduce,past,recent,n;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d",&n);
        for (int i=0;i<n;i++)
        scanf("%s%d%d",homework[i].name, &homework[i].deadline, &homework[i].needtime);
        int maxstate=1<<n;
        for (int s=1;s<maxstate;s++)
        {
            dp[s].score=inf;
            for (int i=n-1;i>=0;i--)
            {
                recent=1<<i;
                if (s & recent)
                {
                    past=s-recent;
                    reduce=dp[past].time+homework[i].needtime-homework[i].deadline;
                    if (reduce<0) reduce=0;
                    if (dp[s].score>dp[past].score+reduce)
                    {
                        dp[s].pre=i;
                        dp[s].last=past;
                        dp[s].time=dp[past].time+homework[i].needtime;
                        dp[s].score=dp[past].score+reduce;
                    }
                }
            }
        }
        stack<int>path;
        recent=maxstate-1;
        while (recent)
        {
            path.push(dp[recent].pre);
            recent=dp[recent].last;
        }
        printf("%d\n",dp[maxstate-1].score);
        while (!path.empty())
        {
            int top=path.top();
            printf("%s\n",homework[top].name);
            path.pop();
        }
    }
    //fclose(stdin);
    return 0;
}

 

 


联系
我们