hdu 1074 Doing Homework (状压dp)

Doing Homework

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

Total Submission(s): 10029    Accepted Submission(s): 4800


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

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.

 

Author

Ignatius.L

第一次写状压。。。

题意：

共有n份作业 

每份作业的名字 截至时间  完成所需时间

超过规定时间我们就要扣一分，这里我们询问最少扣分

dp

状压dp

0代表这个作业尚未被完成 1代表已完成

共有1<<n-1种情况

我们需要dp出全部完成的最小扣分

详细题解看代码注释：

#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<iostream>
#include<stack>
using namespace std;
#define inf 0x3f3f3f3f
string name[20];///名字
int dead[20];///结束时间
int cost[20];///需要花费的时间
struct data
{
int time,pre,now;/// 到达这个状态所需要的
int score;///用来存储得分
} dp[1<<15];///n最大为15 所以1<<15个位置就够用了
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
cin>>n;
for(int i=0; i<n; i++)
{
cin>>name[i]>>dead[i]>>cost[i];
}
memset(dp,0,sizeof(dp));///清空数组
int right = 1<<n;///状态量
for(int i=1; i<right; i++)///从 1 到 n个 1
{
dp[i].score=inf;///先将时间最大化

4000
for(int j=n-1; j>=0; j--)/// 题目要求按照相同的按照题目顺序更新
{
int tmp= 1<<j;
if(tmp&i)///这个状态可以由nex状态添加一个完成
{
int nex=i-tmp;
int spend_time=dp[nex].time+cost[j]-dead[j];///完成这项作业的耗时-规定期限==扣分数
if(spend_time<0)///如果扣分数为负数那么可以确定不扣分 spend_time=0
spend_time=0;
if(spend_time+dp[nex].score<dp[i].score)///数据更新
{
dp[i].score=dp[nex].score+spend_time;///最小分数
dp[i].pre=nex;///前一个状态
dp[i].now=j;///变为当前状态所需要添加的作业编号
dp[i].time=dp[nex].time+cost[j];///已经花费的时间
}
}
}
}
int tmp=right-1;
cout<<dp[tmp].score<<endl;///输出完成所有作业的最小扣分数
string res[20];
int c=0;
while(tmp)///从后往前存储名字。
{
res[c++]=name[dp[tmp].now];
tmp=dp[tmp].pre;
}
for(int i=c-1;i>=0;i--)///正过来输出名字
{
cout<<res[i]<<endl;
}
}
}