题意:

给你一幅图,问你第二大矩形面积是多少。

思路:

直接一行行跑stack求最大矩阵面积的经典算法,不断更新第二大矩形面积,注意第二大矩形可能在第一大矩形里面。

 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr
#include <string>
#include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
#define mem(a,b) memset(a,b,sizeof(a))
#define fo(a,b,c) for(a=b;a<=c;++a)//register int i
#define fr(a,b,c) for(a=b;a>=c;--a)
#define pr printf
#define sc scanf
void swapp(int &a,int &b);
double fabss(double a);
int maxx(int a,int b);
int minn(int a,int b);
int Del_bit_1(int n);
int lowbit(int n);
int abss(int a);
//const long long INF=(1LL<<60);
const double E=2.718281828;
const double PI=acos(-1.0);
const int inf=(<<);
const double ESP=1e-;
const int mod=(int)1e9+;
const int N=(int); struct node
{
int h,l;
};
stack<node> S;
int n,m;
int Nextl[N][N];
char mp[N][N];
//----------------------------------------------------- int main()
{
sc("%d%d",&n,&m);
for(int i=;i<=n;++i)
sc("%s",mp[i]+);
for(int j=;j<=m;++j)
{
int pos=n;
for(int i=n;i>=;--i)
{
if(mp[i][j]=='')
pos=-;
else
{
if(pos==-)
Nextl[i][j]=,pos=i;
else
Nextl[i][j]=pos-i+;
}
}
}
int ans1=,ans2=;
int ll=,rr=;
for(int i=;i<=n;++i)
{
for(int j=;j<=m;++j)
{
node temp;
temp.h=Nextl[i][j];
temp.l=;
if(S.empty())
S.push(temp);
else
{
int L=;
while(!S.empty()&&temp.h<=S.top().h)
{
L+=S.top().l;
int t=L*S.top().h;
if(t>ans1)
{
ans2=ans1;
ll=L,rr=S.top().h;
ans1=t;
}
else
{
if(t>=ans2)
ans2=t;
}
S.pop();
}
temp.l+=L;
S.push(temp);
}
}
int L=;
while(!S.empty())
{
L+=S.top().l;
int t=L*S.top().h;
if(t>ans1)
{
ll=L,rr=S.top().h;
ans2=ans1;
ans1=t;
}
else
{
if(t>=ans2)
ans2=t;
}
S.pop();
}
}
pr("%d\n",maxx(ans2,maxx((ll-)*rr,(rr-)*ll)));
return ;
} /**************************************************************************************/ int maxx(int a,int b)
{
return a>b?a:b;
} void swapp(int &a,int &b)
{
a^=b^=a^=b;
} int lowbit(int n)
{
return n&(-n);
} int Del_bit_1(int n)
{
return n&(n-);
} int abss(int a)
{
return a>?a:-a;
} double fabss(double a)
{
return a>?a:-a;
} int minn(int a,int b)
{
return a<b?a:b;
}
05-08 14:51