2019牛客多校2 H Second Large Rectangle(悬线法)

题意：

求第二大子矩形

思路：

设最大子矩形x*y,第二大子矩形一定在一下情况中

(x-1)*y

x*(y-1)

其他最大子矩形候选者

注意去重手法

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
//#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
     
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
 
using namespace std;
 
typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;
 
const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 1e4+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
//const db pi = acos(-1.0);
 
int a[1111][1111];
int l[1111][1111];
int r[1111][1111];
multiset<int>ans;
int h[1111][1111];
int lft[1111][1111];
int rt[1111][1111];
int n, m;

int main(){
    scanf("%d %d" ,&n, &m);
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            scanf("%1d",&a[i][j]);
        }
    }
    for(int i = 1; i <= n; i++){
        int tmp = 0;
        for(int j = 1; j <= m; j++){
            if(a[i][j]==0)tmp=j;
            lft[i][j]=tmp;
        }
        tmp=m+1;
        for(int j = m; j >= 1; j--){
            if(a[i][j]==0)tmp=j;
            rt[i][j]=tmp;
        }
    }
       PI mx=make_pair(-1,-1);
       int up = -1;
       int mxs = 0;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(i==1||a[i-1][j]==0)h[i][j]=1;
            else h[i][j]=h[i-1][j]+1;
            if(h[i][j]==1){
                l[i][j] = lft[i][j];
                r[i][j] = rt[i][j];
            }
            else{
                l[i][j] = max(l[i-1][j],lft[i][j]);
                r[i][j] = min(r[i-1][j], rt[i][j]);
            }
            if(a[i][j]==0)continue;
            int res = (r[i][j]-l[i][j]-1)*h[i][j];
            if(res>mxs){
                mxs=res;
                mx = make_pair(i,j);up=i-h[i][j]+1;
            }
        }
    }
    int ans = max((r[mx.fst][mx.sc]-l[mx.fst][mx.sc]-1)*(h[mx.fst][mx.sc]-1),(r[mx.fst][mx.sc]-l[mx.fst][mx.sc]-2)*h[mx.fst][mx.sc]);
    //printf("%d
",ans);
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(l[i][j]==l[mx.fst][mx.sc]&&r[i][j]==r[mx.fst][mx.sc]&&i-h[i][j]+1==up)continue;
            int sum = (r[i][j]-l[i][j]-1)*h[i][j];
            if(sum>ans){
                ans=sum;
            }
        }
    }
    printf("%d",ans);
    return 0;
}
/*
1 2
11
3 3
110
111
011
3 3
111
011
011
1 4
1011
3 4
1101
0111
1111
7 8
11110000
11110000
00000111
01110111
01110111
01110000
00000000
4 4
1111
1111
1111
1111
3 3
111
010
010
2 6
010011
001111
3 3
011
111
111
 */