Doing Homework 简单dp&&状态压缩

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

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解释看代码中

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/*
状态压缩枚举状态数，
适合记录状态，
方便在不打乱次序的状态下，
记录最优的状态。
值得学习。
*/
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<vector>
#define inf 0x7fffffff
using namespace std;
int n,m,i,j,k;
struct node
{
    int d,c;
    char ch[101];
}q[1001];
struct stx
{
    int now,pre,time,rs;
}dp[1<<15];
void solve()
{
    int it,jt,kt;
    for(it=1;it<=m;it++)
     dp[it].rs=inf;
     dp[0].time=0;
    for(it=0;it<=m;it++)
     for(jt=0;jt<n;jt++)//二进制 ，枚举每一种状态
     {
       if(!(it&(1<<jt)))//确保不存在重复
       {
           int sta=(it|(1<<jt));//记录状态
           int time=dp[it].time+q[jt].c;
           int rs=max(0,time-q[jt].d);//记录罚时
           rs+=dp[it].rs;
           if(dp[sta].rs>rs)//状态转移
           {
               dp[sta].time=time;
               dp[sta].rs=rs;
               dp[sta].now=jt;
               dp[sta].pre=it;
           }

       }
     }
}
void output(int x)//递归的输出记录的结果
{
    if(x==0)return;
    output(dp[x].pre);
    printf("%s
",q[dp[x].now].ch);
}
int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d",&n);
        m=(1<<n)-1;//算出状态数
        for(i=0;i<n;i++)
        {
            scanf("%s %d %d",q[i].ch,&q[i].d,&q[i].c);
        }
        solve();
        printf("%d
",dp[m].rs);//输出最优结果
        output(m);
    }
    return 0;
}