Description
Input
Output
Sample Input
樣例輸入1:
3 2
4 0
-10 8
-2 -2
Sample Output
樣例輸出1:
4
Data Constraint
The Solution
先n^2枚舉矩形的左右兩邊l,r,然後在行上找最值
即先確定兩邊,再處理行。
我們可以得到前i行l~r的總和。記爲si
我們可以用前綴和來弄。
如果si>sj且i>j則(i-j)*(r-l+1)可以更新答案
維護單調遞減棧
用一個數組f記錄j
每次在其中二分找j
複雜度是
CODE
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define fo(i,a,b) for (int i=a;i<=b;i++)
#define N 405
#define INF 1000000000
using namespace std;
typedef long long ll;
int n,m,top;
ll a[N][N],Stack[N],Pre[N],f[N],ans = 0;
int Find(ll x)
{
int l = 1, r = top,res = -1;
while (l <= r)
{
int mid = (l + r) >> 1;
if (Stack[mid] < x) res = mid , r = mid - 1;
else l = mid + 1;
}
return res;
}
void Work(int x,int y)
{
top = 0;
fo(i,1,n) Pre[i] += a[i][y];
Stack[0] = INF;
ll sum = 0;
fo(i,1,n)
{
sum += Pre[i];
if (sum > 0) ans = max(ans,(ll)i * (y-x+1));
else
{
int t = Find(sum);
if (t != -1) ans = max(ans,(ll)(i-f[t]) * (y-x+1));
}
if (sum < Stack[top]) Stack[++ top] = sum,f[top] = i;
}
}
int main()
{
freopen("max.in","r",stdin);
freopen("max.out","w",stdout);
scanf("%d%d",&n,&m);
fo(i,1,n) fo(j,1,m) scanf("%lld",&a[i][j]);
fo(i,1,m)
{
fill(Pre,Pre+1+N,0);
fo(j,i,m) Work(i,j);
}
printf("%lld\n",ans);
return 0;
}