ZRDay6A. 萌新拆塔(三进制状压dp)

/*
首先打完怪之后吃宝石不一定是最优的
因此我们需要枚举出每个怪物的状态
0：没打 
1：打了怪物 没吃宝石
2：打了怪物 吃了宝石
如果我们能知道打了那些怪，吃了那些宝石，那么此时的状态时确定的
 
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#define int long long 
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
char buf[(1 << 22)], *p1 = buf, *p2 = buf;
#define LL long long 
using namespace std;
const int MAXN = 2 * 1e6 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int T, N;
int lim, Hp, A, F, M, Mp[MAXN], IA[MAXN], IF[MAXN], IM[MAXN];//攻击 血量 防御 魔防 
LL f[MAXN], Po[16];
struct Enemy {
    int H, A, D, S, ap, dp, mp, hp;
}a[15];
vector<int> v[16];
void init() {
    memset(f, 0, sizeof(f));
    Hp = read(); A = read(); F = read(); M = read();
    N = read();    
    lim = Po[N];
    for(int i = 1; i <= N; i++) {
        a[i].H = read();  a[i].A = read(); a[i].D = read(); a[i].S = read(); 
        a[i].ap = read(); a[i].dp = read(); a[i].mp = read(); a[i].hp = read();
    }
    int K = read();
    for(int i = 1; i <= N; i++) v[i].clear();
    for(int i = 1; i <= K; i++) {
        int u = read(), vv = read();
        v[vv].push_back(u);
    }
}
LL Attack(int sta, int id) {
    LL Now = f[sta], A = IA[sta], D = IF[sta], M = IM[sta];
    LL s = a[id].S, h = a[id].H, aa = a[id].A, d = a[id].D;
    if (s & 8) aa = A, d = D; // 模仿 
    if (s & 2) D = 0; // 无视防御 
    LL AA = max(0ll, A - d); // 勇士造成伤害 
    aa = max(0ll, aa - D) * (((s >> 2) & 1) + 1); // 怪物造成的攻击力，是否连击 
    if (AA == 0) return 0; 
    LL t1 = (h - 1) / AA + 1; // 需要打怪几次 
    LL t2 = (s & 1) ? (t1 * aa) : ((t1 - 1) * aa); // 怪造成的攻击力，是否有先攻 
    LL t3 = max(0ll, t2 - M); // 减去魔防 
    return max(0ll, Now - t3);
}
void solve() {
    f[0] = Hp;
    for(int sta = 0; sta < lim; sta++) {
        IA[sta] = A; IF[sta] = F; IM[sta] = M;
        for(int i = 1; i <= N; i++) 
            if(sta / Po[i - 1] % 3 == 2)
                IA[sta] += a[i].ap, IF[sta] += a[i].dp, IM[sta] += a[i].mp;         
    }
    for(int sta = 0; sta < lim; sta++) {
        if(f[sta] == 0) continue;
        for(int i = 1; i <= N; i++) {
            if(sta / Po[i - 1] % 3 == 0) {// not kill 
                bool flag = 0;
                for(int j = 0; j < v[i].size(); j++) 
                    if(sta / Po[v[i][j] - 1] % 3 == 0) // not kiil
                        {flag = 1; break;}
                if(flag == 1) continue;
                LL nhp = Attack(sta, i);
                if(nhp > 0) 
                    f[sta + Po[i - 1]] = max(f[sta + Po[i - 1]], nhp);
            } else if(sta / Po[i - 1] % 3 == 1) {
                f[sta + Po[i - 1]] = max(f[sta + Po[i - 1]], f[sta] + a[i].hp);
            }
        }
    }
    printf("%lld\n", f[lim - 1] == 0 ? -1 : f[lim - 1]);
}
main() {
    Po[0] = 1;
    for(int i = 1; i <= 15; i++) Po[i] = 3 * Po[i - 1];
    T = read();
    while(T--) {
        init();
        solve(); 
    }
    return 0;
}
/*
2 2 1
1 1
2 1 1
*/