Parking Lot CodeForces - 480E

r囧r小猫 2021-11-05 22:50 259阅读 0赞

大意: 给定01矩阵, 单点赋值为1, 求最大全0正方形.

将询问倒序处理, 那么答案一定是递增的, 最多增长$O(n)$次, 对于每次操作暴力判断答案是否增长即可, 也就是说转化为判断是否存在一个边长$x$的正方形包含给定点, 可以维护左右两侧第一个1的位置, 从上往下滑动窗口即可$O(n)$判断, 总复杂度$O(n^2)$

#include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <math.h>
    #include <set>
    #include <map>
    #include <queue>
    #include <string>
    #include <string.h>
    #include <bitset>
    #define REP(i,a,n) for(int i=a;i<=n;++i)
    #define PER(i,a,n) for(int i=n;i>=a;--i)
    #define hr putchar(10)
    #define pb push_back
    #define lc (o<<1)
    #define rc (lc|1)
    #define mid ((l+r)>>1)
    #define ls lc,l,mid
    #define rs rc,mid+1,r
    #define x first
    #define y second
    #define io std::ios::sync_with_stdio(false)
    #define endl '\n'
    #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
    using namespace std;
    typedef long long ll;
    typedef pair<int,int> pii;
    const int P = 1e9+7, INF = 0x3f3f3f3f;
    ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
    ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
    ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
    //head
    
    
    
    
    
    const int N = 2e3+10;
    int n, m, k, ans, Ans[N];
    char s[N][N];
    int dp[N][N], L[N][N], R[N][N], x[N], y[N];
    void upd(int i) {
    	REP(j,1,m) if (s[i][j]=='.') L[i][j] = L[i][j-1]?L[i][j-1]:j;
    	PER(j,1,m) if (s[i][j]=='.') R[i][j] = R[i][j+1]?R[i][j+1]:j;
    }
    void DP() {
    	REP(i,1,n) REP(j,1,m) if (s[i][j]=='.') { 
    		dp[i][j] = min({dp[i-1][j],dp[i][j-1],dp[i-1][j-1]})+1;
    		ans = max(ans,dp[i][j]);
    	}
    	REP(i,1,n) upd(i);
    }
    int chk(int U, int D, int y, int v) {
    	if (D-U+1<v) return 0;
    	deque<int> q1, q2;
    	REP(i,U,D) {
    		if (q1.size()&&i-q1[0]==v) q1.pop_front();
    		if (q2.size()&&i-q2[0]==v) q2.pop_front();
    		while (q1.size()&&L[i][y]>=L[q1.back()][y]) q1.pop_back();
    		while (q2.size()&&R[i][y]<=R[q2.back()][y]) q2.pop_back();
    		q1.push_back(i), q2.push_back(i);
    		if (i-U+1>=v&&R[q2[0]][y]-L[q1[0]][y]+1>=v) return 1;
    	}
    	return 0;
    }
    
    int main() {
    	scanf("%d%d%d", &n, &m, &k);
    	REP(i,1,n) scanf("%s", s[i]+1);
    	REP(i,1,k) { 
    		scanf("%d%d", x+i, y+i);
    		s[x[i]][y[i]] = 'X';
    	}
    	DP();
    	PER(i,1,k) {
    		Ans[i] = ans;
    		s[x[i]][y[i]] = '.', upd(x[i]);
    		int U = x[i], D = x[i];
    		while (s[U][y[i]]=='.') --U; ++U;
    		while (s[D][y[i]]=='.') ++D; --D;
    		while (chk(U,D,y[i],ans+1)) ++ans;
    	}
    	REP(i,1,k) printf("%d\n", Ans[i]);
    }

转载于:https://www.cnblogs.com/uid001/p/10673887.html