hdu1074--Doing Homework（状压dp）

Doing Homework

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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. 

给出了n门作业，每科做也的提交时间和需要时间，每超过一天罚1分，问最少罚几分，要求最小字典序

状态表示，在n科中做了哪几科，二进制数中第i个为1表示做了第i科，dp[i]表示状态为i时的最少罚时，和已经用了的分数

状态转移方程从科目为j的转移到科目为j+1的

为保证字典序最小，所以要让dp[i]尽量有更小的i来得到。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
struct node
{
    char s[120] ;
    int d , c ;
} p[20];
int state[20][7000] , num[20] , c[20] , cnt ;
int dp1[20][7000] , dp2[20][7000] , step[20][7000] ; //dp[i][j][]写i科作业在状态j时，0代表超出的时间，1代表需要的时间。
int find1(int x,int i)
{
    //printf("x == %d i == %d\n", x, i) ;
    int low = 0 , mid , high = num[i]-1 ;
    while( low <= high )
    {
        mid = ( low + high ) ;
        //printf("mid--%d\n", mid) ;
        if( state[i][mid] == x )
            return mid ;
        else if( state[i][mid] > x )
            high = mid - 1 ;
        else
            low = mid + 1 ;
    }
}
int main()
{
    int t , n , i , j , k , l , x , y , temp ;
    scanf("%d", &t) ;
    while( t-- )
    {
        memset(num,0,sizeof(num)) ;
        memset(dp1,-1,sizeof(dp1)) ;
        memset(step,-1,sizeof(step)) ;
        memset(dp2,-1,sizeof(dp2));
        scanf("%d", &n) ;
        for(i = 1 ; i <= n ; i++)
            scanf("%s %d %d", p[i].s, &p[i].d, &p[i].c) ;
        x = 1<<n ;
        for(i = 0 ; i < x ; i++)
        {
            for(j = 0 , temp = 0 ; j < n ; j++)
                if( (1<<j) & i )
                    temp++ ;
            state[temp][ num[temp]++ ]=  i ;
        }
        /*
        for(i = 0 ; i <= n ; i++)
        {
            for(j = 0 ;  j < num[i] ; j++)
                printf("%d ", state[i][j]) ;
            printf("\n") ;
        }*/
        dp1[0][0] = dp2[0][0] = 0 ;
        for(i = 0 ; i < n ; i++)
        {
            for(j = 0 ; j < num[i] ; j++)
            {
                for(k = 1 ; k <= n ; k++)
                {
                    if( 1<<(k-1) & state[i][j] )
                        continue ;
                    l = find1(state[i][j]+(1<<(k-1)),i+1) ;
                    //printf("l == %d\n", l ) ;
                    if( dp1[i+1][l] == -1 || ( dp1[i+1][l] > dp1[i][j] + max(dp2[i][j]+p[k].c-p[k].d,0) ) )
                    {
                        dp1[i+1][l] = dp1[i][j] + max(dp2[i][j]+p[k].c-p[k].d,0) ;
                        dp2[i+1][l] = dp2[i][j]+p[k].c ;
                        step[i+1][l] = j ;
                    }
                }
            }
        }
        /*for(i = 0 ; i <= n ; i++)
        {
            for(j = 0 ; j < num[i] ; j++)
                printf("%d ", dp[i][j][0]) ;
            printf("\n") ;
        }*/
        cnt = 0 ;
        for(i = n , j = 0 ; i > 0 ; i--)
        {
            x = state[i][j] ;
            y = state[i-1][ step[i][j] ] ;
            //printf("x = %d y = %d\n", x, y) ;
            for(k = 1 ; k <= n ; k++)
            {
                if( x % 2 != y % 2 )
                    break ;
                x /= 2 ;
                y /= 2 ;
            }
            c[cnt++] = k ;
            j = step[i][j] ;
        }
        int min1 = dp1[n][0] ;
        printf("%d\n", min1) ;
        for(i = cnt - 1 ; i >= 0 ; i--)
            printf("%s\n", p[c[i]].s ) ;
    }
    return 0 ;
}