HDU 1074 Doing Homework(状态压缩DP)

题意：有n门课，每门课有截止时间和完成所需的时间，如果超过规定时间完成，每超过一天就会扣1分，问怎样安排做作业的顺序才能使得所扣的分最小

思路：二进制表示。

 #include<iostream>
 #include<string>
 #include<algorithm>
 #include<cstdlib>
 #include<cstdio>
 #include<set>
 #include<map>
 #include<vector>
 #include<cstring>
 #include<stack>
 #include<cmath>
 #include<queue>
 #include <bits/stdc++.h>
 using namespace std;
 #define INF 0x3f3f3f3f
 #define ll long long
 #define clc(a,b) memset(a,b,sizeof(a))
 const int MAXN=<<;
 struct node
 {
     int cost;
     int pre;
     int reduced;
 } dp[MAXN];
 bool visited[MAXN];
 struct course
 {
     int deadtime;
     int cost;
     char name[];
 } course[];
 void output(int status)
 {
     int curjob=dp[status].pre^status;
     int curid=;
     curjob>>=;
     while(curjob)
     {
         curid++;
         curjob>>=;
     }
     if(dp[status].pre!=)
     {
         output(dp[status].pre);
     }
     printf("%s\n",course[curid].name);
 }
 int main()
 {
    // freopen("in.txt","r",stdin);
     int T,n;
     int i,j;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         int upper=<<n;
         int dayupper=;
         for(i=; i<n; i++)
         {
             scanf("%s%d%d",&course[i].name,&course[i].deadtime,&course[i].cost);
             dayupper+=course[i].cost;
         }
         memset(visited,false,sizeof(visited));
         dp[].cost=;
         dp[].pre=-;
         dp[].reduced=;
         visited[]=true;
         int work;
         int tupper=upper-;
         for(j=; j<tupper; j++)
         {
             for(work=; work<n; work++)
             {
                 int cur=<<work;
                 if((cur&j)==)
                 {
                     int curtemp=cur|j;
                     int day=dp[j].cost+course[work].cost;
                     dp[curtemp].cost=day;
                     int reduce=day-course[work].deadtime;
                     if(reduce<)reduce=;
                     reduce+=dp[j].reduced;
                     if(visited[curtemp])
                     {
                         if(reduce<dp[curtemp].reduced)
                         {
                             dp[curtemp].reduced=reduce;
                             dp[curtemp].pre=j;
                         }
                     }
                     else
                     {
                         visited[curtemp]=true;
                         dp[curtemp].reduced=reduce;
                         dp[curtemp].pre=j;
                     }
                 }
             }
         }
         printf("%d\n",dp[tupper].reduced);
         output(tupper);
     }
 }