Doing Homework HDU - 1074 (状压DP)

Doing Homework

                                                                                                        Time Limit: 2000/1000 MS (Java/Others)    

                                                                                                        Memory Limit: 65536/32768 K (Java/Others)

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.

 

Author

Ignatius.L

 

1. 首先状压DP涉及位运算， 先补一下位运算；

    a&b 、a|b  ：  a和b写成二进制每一位做与运算，再转成十进制；

dp[ i ]表示状态 i 下有那几个是最优的;

pre[j]记录状态 j 的前一个状态

代码：

 

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f  //1061109567
#define ll long long
using namespace std;
const int MAXN=16;
struct Node
{
    char name[110];
    int D,C;
}node[MAXN];

int dp[1<<MAXN];
int pre[1<<MAXN];
int n;

void output(int status)
{
    if(status==0)return;
    int t=0;
    for(int i=0;i<n;i++)
      if((status&(1<<i))!=0 && (pre[status]&(1<<i))==0)
      {
          t=i;
          break;
      }
    output(pre[status]);
    printf("%s\n",node[t].name);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i = 0;i < n; i++) scanf("%s%d%d",&node[i].name,&node[i].D,&node[i].C);
        memset(dp, INF, sizeof(dp));
        dp[0]=0;
        for(int i=0;i<(1<<n);i++)
        {
            for(int j=0;j<n;j++)
            {
                if(i&(1<<j))continue;
                int s=0;
                for(int k=0;k<n;k++)
                  if(i&(1<<k))
                     s+=node[k].C;
                s+=node[j].C;
                if(s>node[j].D)s=s-node[j].D;
                else s=0;
                if(dp[i|(1<<j)]>dp[i]+s)
                {
                    dp[i|(1<<j)]=dp[i]+s;
                    pre[i|(1<<j)]=i;
                }
            }
        }
        printf("%d\n",dp[(1<<n)-1]);
        output((1<<n)-1);
    }
    return 0;
}