hdu1074Doing Homework(状态压缩，好题)

Doing Homework

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6601    Accepted Submission(s): 2843

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. 本题是不需要知道起点的状态压缩,而且有输出字典序升序的要求 
    
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
int count1[1<<16][16];
struct node{
    int end1,cost;
    char s[101];
}line[16];

struct Node{
    int num;
    int time;
    int pre;
    int count_;
}dp[1<<16];

int main(){
    int t;
    int n;
    scanf("%d",&t);
    while(t--){
        mem0(dp);
        scanf("%d",&n);
        for(int i=0;i<(1<<n);i++)
            dp[i].num=INF;
        for(int i=0;i<n;i++)
            scanf("%s%d%d",line[i].s,&line[i].end1,&line[i].cost);
        dp[0].num=0;
        for(int s=1;s<(1<<n);s++)
            for(int i=n-1;i>=0;i--)
               if(s&(1<<i)){
                    int tmp=s^(1<<i);
                    int x=dp[tmp].time+line[i].cost-line[i].end1;
                    //printf("i is %d s is %d\n",i,s);
                    //printf("%d\n",x);
                    if(x<=0)
                        x=0;
                    if(x+dp[tmp].num<dp[s].num){
                        dp[s].num=x+dp[tmp].num;
                        dp[s].time=dp[tmp].time+line[i].cost;
                        dp[s].pre=tmp;
                        dp[s].count_=i;
                        /*if(s==((1<<n)-1)){
                            printf("%d\n",x);
                            printf("%d\n",dp[s].num);
                        }*/
                    }
                }
        printf("%d\n",dp[(1<<n)-1].num);
        int cnt=0;
        stack<int>s;
        for(int i=(1<<n)-1;cnt<n;i=dp[i].pre){
            s.push(dp[i].count_);
            cnt++;
        }
        while(!s.empty()){
            printf("%s\n",line[s.top()].s);
            s.pop();
        }
    }
    return 0;
}