HDU1074 Doing Homework —— 状压DP

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=1074

 

Doing Homework

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10195    Accepted Submission(s): 4900

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.

 

 

Author

Ignatius.L

 

 

 

 

代码如下：

 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-8;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 15+10;

struct node
{
    string name;
    int deadline, cost;
}subject[MAXN];
int dp[1<<16], id[1<<16], pre[1<<16];
//id[status]表示当前的状态是由添加哪一科而得来的， id[status]的值即subject数组的下标。
//pre[status]表示当前的状态是由哪一个状态转移过来的。

void Print(int s)
{
    if(!s) return;
    Print(pre[s]);  //追溯上一个状态
    cout<< subject[id[s]].name <<endl;
}

int main()
{
    int T, n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        for(int i = 0; i<n; i++)
            cin>>subject[i].name>>subject[i].deadline>>subject[i].cost;

        memset(dp, -1, sizeof(dp));
        dp[0] = 0;
        for(int s = 0; s<(1<<n); s++)
        {
            int time = 0;
            for(int i = 0; i<n; i++)    //计算当前时间
                if(s&(1<<i))
                    time += subject[i].cost;

            for(int i = 0; i<n; i++)
            {
                if(!(s&(1<<i)))
                {
                    int exceeded = time + subject[i].cost - subject[i].deadline;    //计算超出时间
                    if(exceeded<0) exceeded = 0;
                    int tmp = s|(1<<i);
                    if(dp[tmp]==-1 || dp[tmp]>dp[s]+exceeded)   //更新
                    {
                        dp[tmp] = dp[s] + exceeded;
                        id[tmp] = i;
                        pre[tmp] = s;
                    }
                }
            }
        }

        printf("%d\n", dp[(1<<n)-1]);
        Print((1<<n)-1);
    }
}