C. Maximum Subrectangle
time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output
You are given two arrays a
and b of positive integers, with length n and m
respectively.
Let c
be an n×m matrix, where ci,j=ai⋅bj
.
You need to find a subrectangle of the matrix c
such that the sum of its elements is at most x
, and its area (the total number of elements) is the largest possible.
Formally, you need to find the largest number s
such that it is possible to choose integers x1,x2,y1,y2 subject to 1≤x1≤x2≤n, 1≤y1≤y2≤m, (x2−x1+1)×(y2−y1+1)=s, and x2∑i=x1y2∑j=y1ci,j≤x.
Input
The first line contains two integers n
and m (1≤n,m≤2000
).
The second line contains n
integers a1,a2,…,an (1≤ai≤2000
).
The third line contains m
integers b1,b2,…,bm (1≤bi≤2000
).
The fourth line contains a single integer x
(1≤x≤2⋅109
).
Output
If it is possible to choose four integers x1,x2,y1,y2
such that 1≤x1≤x2≤n, 1≤y1≤y2≤m, and ∑x2i=x1∑y2j=y1ci,j≤x, output the largest value of (x2−x1+1)×(y2−y1+1) among all such quadruplets, otherwise output 0
.
Examples
Input
Copy
3 3
1 2 3
1 2 3
9
Output
Copy
4
Input
Copy
5 1
5 4 2 4 5
2
5
Output
Copy
1
给定一个矩阵，求一个子矩阵的sum<=x的面积的最大值；
n,m∈[ 0,2000 ]；
思路：
题目中给了 a,b 两个数组；
其中 c [ i ] [ j ]=a [ i ] * b [ j ]；
我之前的思路：
明显可以预处理出 c[ i ] [ j ]；
接着可以预处理出 二维前缀和；
如果枚举左上点和右下点，复杂度 O(N^4)；
T到飞起23333
很显然；
子矩阵的和就是对应区间的和的乘积；
区间和可以前缀和预处理；
我们要求某个长度和某个宽度的子矩阵，那么我们只需要存储该长度或宽度对应的区间前缀和的min值；
貌似有点绕
那么O(N^2)就可以了；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10
const int N = 1505;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
ll cal(ll x) {
ll ans = 0;
while (x) {
ans += (x % 10);
x /= 10;
}
return ans;
}
ll a[4002], b[4002];
ll suma[4002], sumb[4002];
int n, m;
ll x;
ll SA[4003], SC[4003];
int main()
{
//ios::sync_with_stdio(false);
rdint(n); rdint(m);
for (int i = 1; i <= n; i++) {
rdllt(a[i]); suma[i] = suma[i - 1] + a[i];
}
for (int i = 1; i <= m; i++) {
rdllt(b[i]); sumb[i] = sumb[i - 1] + b[i];
}
rdllt(x);
int ans = 0;
for (int i = 1; i <= max(n, m); i++)SA[i] = SC[i] = x + 1;
for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j++) {
SA[j - i + 1] = min(SA[j - i + 1], suma[j] - suma[i - 1]);
}
}
for (int i = 1; i <= m; i++) {
for (int j = i; j <= m; j++) {
SC[j - i + 1] = min(SC[j - i + 1], sumb[j] - sumb[i - 1]);
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (i*j > ans&&SA[i] * SC[j] <= x)ans = i * j;
}
}
cout << ans << endl;
}