bzoj 5068: 友好的生物

大意: $n$个生物, 每个生物有$K$种属性, 两个生物之间的友好度通过下式计算.

$Friendliness=(\sum\limits_{i=1}^{K-1}C_i \times \text{属性$i$的差别})-C_K \times \text{属性$K$的差别}$

$C$为给定非负数组, 求友好度最大值.

 

 

$\sum |A_i-B_i| = \max\limits_{f_i\in \{0,1\}} \sum f_i(A_i-B_i)$

暴力枚举正负情况去绝对值号. 

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, k, c[N];
struct _ {
	int a[10];
	bool operator < (const _ & rhs) const {
		return a[k] < rhs.a[k];
	}
} a[N];

int main() {
	scanf("%d%d", &n, &k);
	REP(i,1,k) scanf("%d",c+i);
	REP(i,1,n) REP(j,1,k) { 
		scanf("%d", &a[i].a[j]);
		a[i].a[j] *= c[j];
	}
	sort(a+1,a+1+n);
	int mx = (1<<k-1)-1, ans = 0;
	REP(j,0,mx) {
		int mi = 1e9;
		REP(i,1,n) {
			int r = 0;
			REP(ii,0,k-2) {
				r += j>>ii&1?-a[i].a[ii+1]:a[i].a[ii+1];
			}
			r -= a[i].a[k];
			ans = max(ans, r-mi);
			mi = min(mi, r);
		}
	}
	printf("%d\n", ans);
}