原题传送门
01分数规划
∑ a ∑ b < = m i d \frac{\sum a}{\sum b}<=mid ∑b∑a<=mid
∑ a < = m i d ∗ ∑ b \sum a<=mid*\sum b ∑a<=mid∗∑b
w i = a i − m i d ∗ b i w_i=a_i-mid*b_i wi=ai−mid∗bi
问是否存在 ∑ w i < = 0 \sum w_i<=0 ∑wi<=0
先根据 a a a从大到小分组，相同的 a a a一组，组内按照 b b b升序
这样满足严格小于的那个条件
可以 d p i , j dp_{i,j} dpi,j表示前 i i i组中选了 j j j个第一个任务
然后做一个背包
O ( n 3 l o g ) O(n^3log) O(n3log)
Code：
#include <bits/stdc++.h>
#define maxn 55
#define LL long long
using namespace std;
struct data{
LL a, b;
}a[maxn];
LL dp[maxn][maxn];
vector <int> object[maxn];
int n, m;
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c= getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
bool cmp(data x, data y){
return x.a == y.a ? x.b > y.b : x.a > y.a; }
bool check(LL mid){
memset(dp, 0x3f, sizeof(dp));
dp[0][0] = 0;
for (int i = 1, sum = 0; i <= m; ++i){
sum += object[i].size();
for (int j = 0; j <= sum; ++j){
LL x = 0;
for (int k = 0; k <= object[i].size() && k <= j && j - k >= sum - j; ++k){
if (k) x += 1LL * 1000 * a[object[i][k - 1]].a - mid * a[object[i][k - 1]].b;
dp[i][j] = min(dp[i][j], dp[i - 1][j - k] + x);
}
}
}
for (int i = 0; i <= n; ++i)
if (dp[m][i] <= 0) return 1;
return 0;
}
int main(){
n = read();
for (int i = 1; i <= n; ++i) a[i].a = read();
for (int i = 1; i <= n; ++i) a[i].b = read();
sort(a + 1, a + 1 + n, cmp);
a[0].a = a[1].a - 1;
for (int i = 1; i <= n; ++i){
if (a[i].a != a[i - 1].a) ++m;
object[m].push_back(i);
}
LL l = 0, r = 1e11, ans;
while (l <= r){
LL mid = (l + r) >> 1;
if (check(mid)) ans = mid, r = mid - 1; else l = mid + 1;
}
printf("%lld\n", ans);
return 0;
}