JZOJ [4820]. 【NOIP2016提高A组模拟10.15】最大化

Description



Input



Output



Sample Input

样例输入1：

3 2

4 0

-10 8

-2 -2

Sample Output

样例输出1：

4

Data Constraint

The Solution

先n^2枚举矩形的左右两边l,r，然后在行上找最值

即先确定两边，再处理行。

我们可以得到前i行l~r的总和。记为si

我们可以用前缀和来弄。

如果si>sj且i>j则(i-j)*(r-l+1)可以更新答案

维护单调递减栈

用一个数组f记录j

每次在其中二分找j

复杂度是O(n3log)

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

,Stack
,Pre
,f
,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;
}