状态压缩 hdu #10

You are playing CSGO.
There are n Main Weapons and m Secondary Weapons in CSGO. You can
only choose one Main Weapon and one Secondary Weapon. For each weapon,
it has a composite score S.

The higher the composite score of the weapon is, the better for you.

Also each weapon has K performance evaluations x[1], x[2], …, x[K].(range, firing rate, recoil, weight…)

So you shold consider the cooperation of your weapons, you want two
weapons that have big difference in each performance, for example, AWP +
CZ75 is a good choose, and so do AK47 + Desert Eagle.

All in all, you will evaluate your weapons by this formula.(MW for Main Weapon and SW for Secondary Weapon)

Now you have to choose your best Main Weapon & Secondary Weapon and output the maximum evaluation.

InputMultiple query.

On the first line, there is a positive integer T, which describe the number of data. Next there are T groups of data.

for each group, the first line have three positive integers n, m, K.

then, the next n line will describe n Main Weapons, K+1 integers each line S, x[1], x[2], …, x[K]

then, the next m line will describe m Secondary Weapons, K+1 integers each line S, x[1], x[2], …, x[K]

There is a blank line before each groups of data.

T<=100, n<=100000, m<=100000, K<=5, 0<=S<=1e9, |x[i]|<=1e9, sum of (n+m)<=300000

OutputYour output should include T lines, for each line, output the maximum evaluation for the corresponding datum.Sample Input

2
2 2 1
0 233
0 666
0 123
0 456
2 2 1
100 0 1000 100 1000 100
100 0

Sample Output

543
2000
题意 ： 有 n 种主武器， m 种副武器， 同时每种武器都有 k 个权值，询问上面所给的目标式子中的最大收益
思路分析 ： 考虑一下绝对值的性质， a-b 的绝对值等于 a-b 或者 -a+b , 并且题目所给的 k <= 5, 显然这里我们可以二进制去枚举，记录最大值即可
代码示例：

#define ll long long
const ll maxn = 1e5+5;

ll n, m, k;
ll a[maxn][10], b[maxn][10];
ll sa[50], sb[50];

void init() {
    ll f = 1;
    for(ll i = 1; i <= k; i++) f *= 2;
    memset(sa, 0x8f, sizeof(sa)); memset(sb, 0x8f, sizeof(sb));
    //printf("++ %lld \n", sa[0]);
    for(ll i = 1; i <= n; i++){
        for(ll state = 0; state < f; state++){
            ll sum = 0;
            for(ll j = 0; j < k; j++){
                if (state & (1<<j)) sum += a[i][j+1];
                else sum -= a[i][j+1];
            }
            sa[state] = max(sa[state], sum+a[i][0]);
        }
    }

    for(ll i = 1; i <= m; i++){
        for(ll state = 0; state < f; state++){
            ll sum = 0;
            for(ll j = 0; j < k; j++){
                if (state & (1<<j)) sum += b[i][j+1];
                else sum -= b[i][j+1];
            }
            sb[state] = max(sb[state], sum+b[i][0]);
        }
    }
}

void solve() {
    ll num = 1<<k;
    ll ans = 0x8f;
    for(ll i = 0; i < num; i++){
        ll pp = num-1-i;

        ans = max(ans, sa[i]+sb[pp]);
    }
    printf("%lld\n", ans);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ll t;

    cin >> t;
    while(t--){
        scanf("%lld%lld%lld", &n, &m, &k);
        for(ll i = 1; i <= n; i++){
            for(ll j = 0; j <= k; j++){
                scanf("%lld", &a[i][j]);
            }
        }
        for(ll i = 1; i <= m; i++){
            for(ll j = 0; j <= k; j++){
                scanf("%lld", &b[i][j]);
            }
        }
        init();
        solve();
    }
    return 0;
}