hdu 2809(状压dp）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2809

思路：简单的状压dp,看代码会更明白。

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 using namespace std;

 6 

 7 struct State{

 8     int ati,def,hp,lev,exp;

 9 }dp[1<<20+2];

10 

11 struct Node{

12     int ati,def,hp,exp;

13 }node[21];

14 

15 int n;

16 int In_ati,In_def,In_hp;

17 char str[22];

18 

19 int main()

20 {

21     while(~scanf("%d%d%d%d%d%d",&dp[0].ati,&dp[0].def,&dp[0].hp,&In_ati,&In_def,&In_hp)){

22         scanf("%d",&n);

23         for(int i=0;i<n;i++){

24             scanf("%s%d%d%d%d",str,&node[i].ati,&node[i].def,&node[i].hp,&node[i].exp);

25         }

26         for(int i=1;i<(1<<20)+2;i++){

27             dp[i].hp=0;

28         }

29         dp[0].lev=1;

30         dp[0].exp=0;

31         for(int state=0;state<(1<<n);state++){

32             if(dp[state].hp<=0)continue;

33             for(int i=0;i<n;i++){

34                 if(state&(1<<i))continue;

35                 State S=dp[state];

36                 int tmp1=max(1,S.ati-node[i].def);

37                 int tmp2=max(1,node[i].ati-S.def);

38                 int t=(node[i].hp/tmp1)+((node[i].hp%tmp1==0)?-1:0);

39                 S.hp-=t*tmp2;

40                 if(S.hp<=0)continue;

41                 S.exp+=node[i].exp;

42                 if(S.exp>=S.lev*100){

43                     S.lev++;

44                     S.ati+=In_ati;

45                     S.def+=In_def;

46                     S.hp+=In_hp;

47                 }

48                 if(S.hp>dp[state|(1<<i)].hp){

49                     dp[state|(1<<i)]=S;

50                 }

51             }

52         }

53         if(dp[(1<<n)-1].hp<=0){

54             puts("Poor LvBu,his period was gone.");

55         }else 

56             printf("%d\n",dp[(1<<n)-1].hp);

57     }

58     return 0;

59 }

60 

61 

62                 

63 

64                 

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