題意：

求第二大子矩形

思路：

設最大子矩形x*y,第二大子矩形一定在一下情況中

(x-1)*y

x*(y-1)

其他最大子矩形候選者

注意去重手法

代碼：

#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|1using 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\n",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*/

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轉載于:https://www.cnblogs.com/wrjlinkkkkkk/p/11260147.html

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