hdu 1074 Doing Homework

Doing Homework

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6299    Accepted Submission(s): 2708

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, bothComputer->English->Math and Computer->Math->English leads to reduce3 points, but the

word "English" appears earlierthan the word "Math", so we choose the first order. That is so-calledalphabet order.

这题的大意是一个人需要完成他的作业，但是他的老师给他定了期限，每超过期限一天，就扣一分，问你如何安排做作业的顺序，才能使扣的分数最少，输出要扣的最小分数，并输出完成作业的顺序。（科目数目n<=15）

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100
struct node{
    int cost;
    int time;
    int pre;
    int now;
}dp[1<<15+1];
char s[20][N];
int need[N],close[N];
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        for(int i = 0 ; i < n ; i++)
            scanf("%s%d%d",s[i],&close[i],&need[i]);

        memset(dp,0,sizeof(dp));

        for(int i = 1 ; i < (1 << n) ; i++)    {
            dp[i].cost=INF;
            for(int j = n - 1 ; j >= 0 ; j--)    {
                int k=(1 << j);
                if( i & k)   {//查看第j个位置是否要填上
                    int st=dp[i-k].time+need[j]-close[j];
                    if(st<0)
                        st=0;
                    if(st+dp[i-k].cost<dp[i].cost) {
                        dp[i].cost=st+dp[i-k].cost;
                        dp[i].time=dp[i-k].time+need[j];
                        dp[i].now=j;//记录当前位置
                        dp[i].pre=i-k;//记录前面的位置为了递推
                    }
                }
            }
        }

        stack <int> S;
        int t = (1 << n) - 1;
        //for(int i=0;i<t;i++)
        printf("%d\n",dp[t].cost);
        while(t) {
            S.push(dp[t].now);
            t=dp[t].pre;
        }
        while(!S.empty()) {
            printf("%s\n",s[S.top()]);
            S.pop();
        }
    }
    return 0;
}