状压DP。我认为是数据水了,用打死了哪几只作为状态,AC代码只需要保存当前状态的最大血量,完全没有考虑攻击力大小。
个人认为正确DP应该这样的:dp[状态][等级],但这样写不能AC,时间复杂度会很大,但答案应该是正确的。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std; struct X
{
int HP;
int ATI;
int DEF;
int e;
int lv;
} dp[( << ) + ], A[]; int In_ATI, In_DEF, In_HP;
int ATI, DEF, HP;
int n;
int base[]; int MAX(int a, int b)
{
if (a>b) return a;
return b;
} void read()
{
scanf("%d", &n);
for (int i = ; i<n; i++)
{
string name;
cin >> name;
scanf("%d%d%d%d", &A[i].ATI, &A[i].DEF, &A[i].HP, &A[i].e);
}
} void init()
{
dp[].ATI = ATI, dp[].DEF = DEF, dp[].HP = HP;
dp[].e = , dp[].lv = ; for (int i = ; i<( << n); i++) dp[i].HP = -;
} int f(int num1, int num2)
{
int A1 = MAX(, dp[num1].ATI - A[num2].DEF);
int A2 = MAX(, A[num2].ATI - dp[num1].DEF); int HP1 = dp[num1].HP;
int HP2 = A[num2].HP; int tot1, tot2;
if (HP2%A1 == ) tot1 = HP2 / A1;
else tot1 = HP2 / A1 + ; if (HP1%A2 == ) tot2 = HP1 / A2;
else tot2 = HP1 / A2 + ; if (tot1 <= tot2) return HP1 - A2*(tot1 - ); //返回吕布打完后的血量
return -; //返回吕布被打死
} void work()
{
for (int i = ; i<( << n); i++)
{
int tot = , tmp = i;
while (tmp) base[tot++] = tmp % , tmp = tmp / ; for (int j = ; j<tot; j++)
{
if (base[j])
{
int pre = i - ( << j);
if (dp[pre].HP == -) continue; int lx = f(pre, j);
if (lx == -) continue; int hp = lx;
int e = dp[pre].e + A[j].e;
int lv = dp[pre].lv;
int ati = dp[pre].ATI;
int def = dp[pre].DEF; if (e >= * dp[pre].lv)
{
hp = hp + In_HP;
ati = ati + In_ATI;
def = def + In_DEF;
lv++;
} if (hp>dp[i].HP)
{
dp[i].e = e;
dp[i].lv = lv;
dp[i].HP = hp;
dp[i].ATI = ati;
dp[i].DEF = def;
}
}
}
} if (dp[( << n) - ].HP == -) printf("Poor LvBu,his period was gone.\n");
else printf("%d\n", dp[( << n) - ].HP);
} int main()
{
while (~scanf("%d%d%d%d%d%d", &ATI, &DEF, &HP, &In_ATI, &In_DEF, &In_HP))
{
read();
init();
work();
}
return ;
}