每日三题-Day2-A（HDU 1074 Doing Homework 状压DP）

原题地址

Doing Homework

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8928 Accepted Submission(s): 4181

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

23Computer 3 3English 20 1Math 3 23Computer 3 3English 6 3Math 6 3

Sample Output

2ComputerMathEnglish3ComputerEnglishMath
Hint
In the second test case, both Computer->English->Math and Computer->Math->English leads to reduce 3 points, but the word "English" appears earlier than the word "Math", so we choose the first order. That is so-called alphabet order.
 

题目大意：

给你n个作业，以及完成它所需时间和最晚昨晚时间。每门作业如果晚交一天扣一分，问如何安排做作业顺序使得所扣分数最少。如果最少分数有多种方案，字典序小的科目优先。

输出最少扣分，然后按做作业顺序输出科目名字。

注意，科目是按字典序输入的。

解题思路：

n<15，只要对完成科目的顺序全排列一下，找到扣分最小的便可。

使用状态压缩DP可快速求解。

AC代码：

#include<cstdio>#include<algorithm>using namespace std;#define N 1<<15#define INF 0x3f3f3f3fstruct DP{    int cost;    int pre;    int reduce;    int w;    int flag;} dp[N];struct homework{    int name[111];    int time;    int deadline;} hw[20];void dfs(int a,int n){    if(dp[a].pre==-1) return ;    else dfs(dp[a].pre,n);    printf("%s\n",hw[dp[a].w].name);    return ;}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0; i<n; i++)        {            scanf("%s%d%d",hw[i].name,&hw[i].deadline,&hw[i].time);        }        dp[0].cost=0;        dp[0].reduce=0;        dp[0].pre=-1;        for(int i=1; i<(1<<n); i++)        {            dp[i].flag=0;        }        for(int i=0; i<(1<<n); i++)        {            for(int j=0; j<n; j++)            {                if(!(i&(1<<j)))                {                    int sta = i|(1<<j);                    int tmp=dp[i].reduce+max(dp[i].cost+hw[j].time-hw[j].deadline,0);                    if(dp[sta].flag==1)                    {                        if(dp[sta].reduce>tmp)                        {                            dp[sta].cost=dp[i].cost+hw[j].time;                            dp[sta].pre = i;                            dp[sta].w = j;                            dp[sta].reduce = tmp;                        }                    }                    else                    {                        dp[sta].cost=dp[i].cost+hw[j].time;                        dp[sta].pre = i;                        dp[sta].w = j;                        dp[sta].reduce = tmp;                        dp[sta].flag=1;                    }                }            }        }        int p=(1<<n)-1;        printf("%d\n",dp[p].reduce);        /*        for(int i=0;i<=7;i++)        printf("%d %d %d %d %d\n",dp[i].cost,dp[i].reduce,dp[i].pre,dp[i].w,dp[i].flag);        */        int s=0;        dfs((1<<n)-1,n);    }    return 0;}