HDU1074：Doing Homework-状态压缩DP

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.
 

题意：有n份作业，每份作业有对应的截止时间和花费时间，求做完所有作业至少逾期多少天；

思路：非常像贪心，但是没有贪心思路，最多只有15种作业，所以可以考虑状态压缩；

AC code：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <string>
#include <map>
using namespace std;
#define LL long long
#define INF 0x3f3f3f3f
#define fi first
#define se second
#define eps 1
const int mod=1000000000+7;
const int maxn=16;

struct node{
    string name;
    int deadline,cost;
}hw[maxn];
int dp[1<<maxn];
int pre[1<<maxn];

void print(int x,int n){
    if(x==0) return;
    //递归打印
    print(pre[x],n);
    for(int i=0;i<n;i++){
        //当前状态为1，前驱状态为0即为更新的状态
        if(((x>>i)&1)==1&&((pre[x]>>i)&1)==0){
            cout<<hw[i].name<<endl;
            break;
        }
    }
}

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


        memset(dp,INF,sizeof dp);
        memset(pre,-1,sizeof pre);
        int ed=(1<<n)-1;    //终点当然是全1状态
        dp[0]=0;            //什么都不选即为0
        for(int i=0;i<=ed;i++){
            int sum=0;
            //sum为每种状态的花费时间总和
            for(int j=0;j<n;j++){
                if((i>>j)&1) sum+=hw[j].cost;
            }
            for(int j=0;j<n;j++){
                //找可以更新的点
                if(((i>>j)&1)==0){
                    int time=sum+hw[j].cost;
                    //time为更新后所耗费的时间
                    if(time>hw[j].deadline) time-=hw[j].deadline;
                    else time=0;
                    //从前往后递推
                    if(dp[i|(1<<j)]>dp[i]+time){
                        dp[i|(1<<j)]=dp[i]+time;
                        pre[i|(1<<j)]=i;
                    }
                }
            }
        }
        printf("%d\n",dp[ed]);
        print(ed,n);
    }
    return 0;
}