二进制状态压缩
abs(a - b) = max(a - b, b - a)
通过上式我们可以发现，对于选中的武器的每个属性，主武器如果是+，那么服务器必定是-，反之也是一样。
我们对所有绝对值求和其实就是把每个数前面带上符号相加。比如abs(a-b)+abs(c-d),(a-b>b-a, d-c>c-d) = a-b+d-c = a-c-b+d
所以我们可以枚举所有武器的每种属性的符号，求出他们和的最大值，最后在把副武器状态取反相加就可以了。
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
int ret = 0, w = 0; char ch = 0;
while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();
return w ? -ret : ret;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template <typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template <typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
A ans = 1;
for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
return ans;
}
const int N = 100005;
int _, n, m, k, sc[N];
ll mw[40], sw[40];
int main(){
for(_ = read(); _; _ --){
for(int i = 0; i < 40; i ++) mw[i] = sw[i] = -INF;
n = read(), m = read(), k = read();
for(int i = 1; i <= n; i ++){
int basic = read();
for(int j = 1; j <= k; j ++) sc[j] = read();
for(int j = 0; j < (1 << k); j ++){
ll val = basic;
for(int t = 0; t < k; t ++){
if((j >> t) & 1) val += sc[t + 1];
else val -= sc[t + 1];
}
mw[j] = max(mw[j], val);
}
}
for(int i = 1; i <= m; i ++){
int basic = read();
for(int j = 1; j <= k; j ++) sc[j] = read();
for(int j = 0; j < (1 << k); j ++){
ll val = basic;
for(int t = 0; t < k; t ++){
if((j >> t) & 1) val += sc[t + 1];
else val -= sc[t + 1];
}
sw[j] = max(sw[j], val);
}
}
ll ans = -INF;
for(int i = 0; i < (1 << k); i ++){
ans = max(ans, mw[i] + sw[(1 << k) - 1 - i]);
}
printf("%lld\n", ans);
}
return 0;
}