God of War Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit  Status

Description

At 184~280 A.D ,there were many kingdoms in China. Three strongest among them are "Wei", "Shu", "Wu". People call this period as "Three Kingdoms". 
HH is a super "Three Kingdoms" fan, because at this period there were many heroes and exciting stories. Among the heroes HH worships LvBu most. 
LvBu is the God of War and is also intelligent, but his ambition is too big while enemies are too powerful .Many monarchs wanted to kill him. 
At 198 A.D ,CaoCao fought with LvBu at Xuzhou.Though Lvbu is the God of War ,CaoCao had so many generals: Xuchu,DianWei XiahouChun……Facing so many heroes ,could LvBu beat all of them? 
 
Given the LvBu's ATI, DEF, HP, and enemies’ ATI, DEF,HP, experience (if LvBu killed one of his enemies, he can get that experience ,and if his experience got more than or equal to 100*level,he would level-up and become stronger) and the In_ATI,In_DEF,In_HP(indicating when LvBu levels up,his ability will increase this point). 
Each turn LvBu will choose an enemy to fight. Please help LvBu find a way to beat all of enemies and survive with the max HP. 
Here’s a fight between LvBu and A: 
If LvBu attack A, A will lose Max(1,LvBu's ATI- A's DEF) hp;
If A survived, he will give LvBu Max(1,A'ATI- LvBu'DEF) injury.
If LvBu is still alive, repeat it untill someone is dead(hp <= 0). 
LvBu's initial level is 1 and experience is 0,and he can level up many times.

Input

The input contains at most 20 test cases. 
For each case , the first line contains six intergers ,indicating LvBu's ATI,DEF,HP and In_ATI,In_DEF,In_HP. 
The next line gives an interger N(0<N<=20),indicating the number of the enemies . 
Then N lines followed, every line contains the name(the length of each name is no more than 20),ATI,DEF,HP, experience(1<experience<=100).

Output

If LvBu is dead output "Poor LvBu,his period was gone." 
Or output the maximum HP left.

Sample Input

100  80  100  5  5  5
2
ZhangFei 95  75  100  100 
XuChu 90  90  100  90

100 75 100 5 5 5
1
GuanYu 95 85 100 100

Sample Output

30
Poor LvBu,his period was gone.

題目大意:奧特曼很牛逼，要單挑n只怪獸。怪獸和奧特曼一樣都有hp、攻擊力、防禦力，奧特曼有一個經驗值屬性，通過打怪獸獲得經驗值超過100就升級，升級時hp加一些，攻擊力加一些，防禦力加一些，回不到滿血狀態，奧特曼每次都要和怪獸血拼，奧特曼先打，怪獸後打，直到一方倒下為止。問奧特曼能不能打倒所有的怪獸從而拯救地球？可以的話輸出剩下的最大hp。

解題思路：題目看起來很雷人，實際上就是普通的TSP，只是每次狀態轉移略顯麻煩些。為什麼可以轉換為TSP呢？因為每隻怪獸都只要打一次，而且打完若干只怪獸後剩下的經驗值、攻擊力、防禦力都是一樣的，只有hp會隨順序改變，那麼用一個數組dp[i]表示i狀態時的最多hp，每次更新都去計算i狀態下的各項屬性。我把hp,at,def,ex都封裝進一個結構體，意思是一樣的，狀態轉移見程式碼。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
const int N = (1<<20)+10;
int n;
struct hero
{
    int ati,def,hp,ex, lev;
}dp[N];
struct enemy
{
    int ati,def,hp,ex;
}e[N];
void init()
{
    for(int i=1;i<(1<<n);i++)
        dp[i].ati=dp[i].def=dp[i].ex=dp[i].hp=0;
    return ;
}
char str[25];


int main()
{
    int ati,def,hp;
    while(scanf("%d %d %d %d %d %d",&dp[0].ati, &dp[0].def,&dp[0].hp, &ati, &def, &hp)!=EOF)
    {
        scanf("%d",&n);
        init();
        for(int i=0;i<n;i++)
            scanf("%s %d %d %d %d",str,&e[i].ati,&e[i].def,&e[i].hp,&e[i].ex);
        dp[0].ex=0, dp[0].lev=1;
        for(int i=0;i<(1<<n);i++)
        {
            if(dp[i].hp<=0)
                continue;
            for(int j=0;j<n;j++)
            {
                if(i&(1<<j))
                    continue;
                hero tmp=dp[i];
                int att1=max(1,tmp.ati-e[j].def);
                int att2=max(1,e[j].ati-tmp.def);
                int time=e[j].hp/att1+(e[j].hp%att1==0?-1:0);
                tmp.hp-=(att2*time);
                if(tmp.hp<=0)
                    continue;
                int state=i|(1<<j);
                tmp.ex+=e[j].ex;
                if(tmp.ex>=tmp.lev*100)
                    tmp.ati+=ati,tmp.def+=def,tmp.hp+=hp,tmp.lev++;
                if(dp[state].hp<tmp.hp)
                {
                    dp[state]=tmp;
                }
            }
        }
        if(dp[(1<<n)-1].hp<=0)
            printf("Poor LvBu,his period was gone.\n");
        else
            printf("%d\n",dp[(1<<n)-1].hp);
    }
    return 0;
}