HUD 1074 doing homework(状压DP）

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状态压缩的一道题，刚开始我根本想不到是状态压缩，我以为会用什么贪心算法做出来，看了网上大神的解法才知道是要位压缩，感觉不可思议。

状态变量s从0—>（1<<n-1)，然后再加上一个循环变量for（i：1->n),二进制的s里面第j位为1表示第j个作业已经做完了，然后对做完第j个作业的之后的状态进行刷新；换一种说法就是刷新s状态是根据能达到这一状态做的最后一种作业来定的；

Doing Homework

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8124 Accepted Submission(s): 3733

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

#include<stdio.h>#include<string.h>#include<math.h>#include <iostream>#include <algorithm>using namespace std;const int MAXNUM=1<<15;const int INF = 0x7fffffff;struct{    char name[101];    int deadline;    int workday;}homework[20];struct{    int reduce;    int cost;    int piror;  //该状态之前的最后写的一们作业；}dp[MAXNUM+100];int cases,n;int visited[MAXNUM+100];bool  loop(){    if(!cases)        return false;    cases--;    memset(visited,0,sizeof(visited));    cin>>n;    for(int i=1;i<=n;i++)    {        cin>>homework[i].name>>homework[i].deadline>>homework[i].workday;    }    long long maxstation=(1<<n)-1;    for(long s=0;s<maxstation;s++)    {          for(int i=1;i<=n;i++)          {              if((s&(1<<(i-1)))==0)              {                  long long thistation=s+(1<<(i-1));                  int thiswork=dp[s].cost+homework[i].workday;                  int thisreduce=thiswork-homework[i].deadline;                  if(thisreduce<0)                    thisreduce=0;                  if((visited[thistation]&&dp[s].reduce+thisreduce<dp[thistation].reduce))                  {                      dp[thistation].reduce=dp[s].reduce+thisreduce;                      dp[thistation].piror=i;                      dp[thistation].cost=thiswork;                  }                  if((visited[thistation]==0))                  {                       dp[thistation].reduce=dp[s].reduce+thisreduce;                      dp[thistation].piror=i;                      dp[thistation].cost=thiswork;                      visited[thistation]=1;                  }              }          }    }    long long total=(1<<n)-1;    cout<<dp[total].reduce<<endl;    int j=n;    int outs[n+1];    while(total)    {        int k=dp[total].piror;        outs[j]=k;        j--;        total-=(1<<(k-1));    }    for(int i=1;i<=n;i++)    {        cout<<homework[outs[i]].name<<endl;    }    return true;}int main(){    cin.tie(0);    cin.sync_with_stdio(false);  // freopen("in.txt","r",stdin);    cin>>cases;    while(loop());    return 0;}

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