hdu 1074

For state compression questions, it should be noted that there are many situations where the deduction is the same, so the output should be in lexicographical order, that is, the input should come first, and the output should also come first. See code comments for details.
Reference article:

• http://www.voidcn.com/article/p-fsgosnqv-bpb.html
• http://www.cnblogs.com/kuangbin/archive/2011/08/04/2127687.html
• http://www.cnblogs.com/qkhhxkj/archive/2011/06/29/2093894.html

---

#include <iostream>
#include <stdio.h>
#include <string>
#include <stack>
using namespace std;

const int maxn=1<<17;
struct homework{
    char name[150];
    int deadline;
    int timecost;//做作业所花费的时间
}h[30];
struct node{
    int score;//当前的扣分情况
    int time;//当前所花的天数
    int num;//当前所做作业的序号
    int pre;//上一个状态的地址
}dp[maxn];//数组要开的足够大,数字上限为2^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",&h[i].name,&h[i].deadline,&h[i].timecost);
        dp[0].score=0;
        dp[0].time=0;
        int M=(1<<N)-1;
        for(int i=1;i<=M;i++){//用二进制的0～n-1位来表示作业1～n的完成情况，0为都没有完成，(1<<N)-1表示所有作业都完成，这里进行遍历
            dp[i].score=100000;//保证dp[i]得到初始化
            for(int j=N-1;j>=0;j--){//j从大到小每个作业遍历，保证字典序输出
                int temp=1<<j;
                if((i&temp)==temp){//表示当前作业状态是做了j号作业的
                    int k=i-temp;//表示上一个状态，也就是没有做j号作业,其余与这个状态一样
                    int x=dp[k].time+h[j].timecost-h[j].deadline;
                    if(x<0)
                        x=0;//计算做j号作业是否逾期，若没有预期逾期则不扣分
                    if(dp[k].score+x<dp[i].score){//若为扣分最少的情况，则更改信息
                        dp[i].score=dp[k].score+x;
                        dp[i].num=j;
                        dp[i].time=dp[k].time+h[j].timecost;
                        dp[i].pre=k;
                    }
                }
            }
        }
        printf("%d\n",dp[M].score);
        stack<int>st;
        while(M!=0){
            st.push(dp[M].num);
            M=dp[M].pre;
        }
        while(!st.empty()){
            printf("%s\n",h[st.top()].name);
            st.pop();
        }

    }
    return 0;
}

I really don't understand this very well, sorry for the mistakes.