状态压缩-----HDU1074 Doing Homework

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 
 6 using namespace std;
 7 #define MAX 1<<15+1
 8 int N;
 9 struct Node{
10 char name[108];
11 int deadline;
12 int cost;
13 }homework[16];
14 struct START
15 {
16     int time;
17     int pre;
18     int score;
19     int now;
20 } dp[MAX];
21 int record[16];
22 
23 void DP()
24 {
25     int Final=(1<<N)-1,punish,previous;
26     dp[0].id=dp[0].now=dp[0].pre=dp[0].score=dp[0].time=0;
27     for(int i=1;i<=Final;i++)
28     {
29         dp[i].score=10000000;
30         for(int j=0;j<N;j++)
31             if(i&(1<<j))
32         {
33              previous=(i-(1<<j));
34             if(dp[previous].time+homework[j].cost>homework[j].deadline)
35                 punish=dp[previous].time+homework[j].cost-homework[j].deadline;
36             else punish=0;
37             if(dp[i].score>=dp[previous].score+punish)
38             {
39                 dp[i].score=dp[previous].score+punish;
40                 dp[i].time =dp[previous].time+homework[j].cost;
41                 dp[i].pre=previous;
42                 dp[i].now=j;
43             }
44         }
45     }
46     printf("%d\n",dp[Final].score);
47     int cal=Final,num=0;
48     while(cal)
49     {
50         record[num++]=dp[cal].now;
51         cal=dp[cal].pre;
52     }
53     for(int i=num-1;i>=0;i--)
54     puts(homework[record[i]].name);
55 
56 }
57 int main()
58 {
59    int T;
60    scanf("%d",&T);
61    while(T--)
62    {
63        scanf("%d",&N);
64        for(int i=0;i<N;i++)scanf("%s%d%d",homework[i].name,&homework[i].deadline,&homework[i].cost);
65        DP();
66    }
67     return 0;
68 }

 1 void DP()
 2 {
 3     int Final=(1<<N)-1,punish,previous;
 4     dp[0].id=dp[0].now=dp[0].pre=dp[0].score=dp[0].time=0;
 5     for(int i=1;i<=Final;i++)//状态从全0到全1
 6     {
 7         dp[i].score=10000000;//为了找出最小值所以先初始化成比较大的值
 8         for(int j=0;j<N;j++)//题编号
 9             if(i&(1<<j))//状态i中有题j,为什么这样写是因为保证在状态全一的时候
10                 //枚举过每一种可能，即由状态previous转化成状态j
11         {
12              previous=(i-(1<<j));//从previous到i仅相差一个作业
13              //若从状态previous再做一题总时间加起来超过了作业j的deadline，记录扣几分
14              //否则为零
15             if(dp[previous].time+homework[j].cost>homework[j].deadline)
16                 punish=dp[previous].time+homework[j].cost-homework[j].deadline;
17             else punish=0;
18             //寻找状态为i时候的最小值
19             if(dp[i].score>=dp[previous].score+punish)
20             {
21                 dp[i].score=dp[previous].score+punish;
22                 dp[i].time =dp[previous].time+homework[j].cost;
23                 dp[i].pre=previous;
24                 dp[i].now=j;
25             }
26         }
27     }
28     printf("%d\n",dp[Final].score);
29     //显示路径now代表的是状态i是由写了一道编号为now的题形成的，previous是i的前一状态
30     int cal=Final,num=0;
31     while(cal)
32     {
33         record[num++]=dp[cal].now;
34         cal=dp[cal].pre;
35     }
36     for(int i=num-1;i>=0;i--)
37     puts(homework[record[i]].name);
38 
39 }