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
