题目大意:给定$n$个$A$类元素和$m$个$B$类元素,每类元素有值$S$和$k$个值$x_0,x_1,dots,x_{k-1}(kleqslant 5)$。
要求选出一个$A$类元素$a$和$B$类元素$b$,最大化$a.S+b.S+sumlimits_{i=0}^{k-1}|a.x_i-b.x_i|$。
题解:发现$kleqslant 5$,可以搞事情,$|a-b|=max(a-b,b-a)$,枚举每个元素的$x_i$是正是负,对每个一个状态求出在$A$和$B$中的最大值,取最大值就行了
卡点:无
C++ Code:
#include <cstdio> #include <cstring> #define maxn 100010 const long long inf = 0x3f3f3f3f3f3f3f3f; int Tim; int n, m, K, U; long long MW[1 << 5 | 3], SW[1 << 5 | 3], x[10]; long long ans; inline long long max(long long a, long long b) {return a > b ? a : b;} void read(int n, long long *W) { long long tmp; for (int i = 1; i <= n; i++) { scanf("%lld", &tmp); for (int j = 0; j < K; j++) scanf("%lld", x + j); for (int j = 0; j < U; j++) { long long tmp6 = tmp; for (int k = 0; k < K; k++) tmp6 += (j & (1 << k)) ? x[k] : -x[k]; W[j] = max(W[j], tmp6); } } } int main() { scanf("%d", &Tim); while (Tim --> 0) { scanf("%d%d%d", &n, &m, &K); ans = -inf; U = 1 << K; for (int i = 0; i < U; i++) MW[i] = SW[i] = -inf; read(n, MW); read(m, SW); for (int i = 0; i < U; i++) { ans = max(ans, MW[i] + SW[(U - 1) ^ i]); } printf("%lld ", ans); } return 0; }