CF 480 E. Parking Lot

CF 480 E. Parking Lot

http://codeforces.com/contest/480/problem/E

题意：

　　给一个n*m的01矩阵，每次可以将一个0修改为1，求最大全0的矩阵。

分析：

　　将询问离线，从后往前处理询问，相当于每次将一个1变成0，答案是递增的。

　　用悬线法或者单调栈来求。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cctype>
#include<set>
#include<vector>
#include<queue>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 2010;

char a[N][N];
int x[N], y[N], L[N], R[N], U[N][N], D[N][N], q1[N], q2[N], ans[N], sk[N];
int Ans, n, m;

void solve() {
    for (int i=1; i<=n; ++i) 
        for (int j=1; j<=m; ++j) 
            U[i][j] = a[i][j] == '.' ? U[i-1][j] + 1 : 0;
    for (int i=n; i>=1; --i) 
        for (int j=1; j<=m; ++j) 
            D[i][j] = a[i][j] == '.' ? D[i+1][j] + 1 : 0;
    for (int i=1; i<=n; ++i) { // 单调栈 
        U[i][0] = U[i][m+1] = -1;
        for (int top=0,j=1; j<=m+1; ++j) {
            while (top > 0 && U[i][j] < U[i][sk[top]]) R[sk[top]] = j, top--;
            sk[++top] = j;
        }
        for (int top=0,j=m; j>=0; --j) {
            while (top > 0 && U[i][j] < U[i][sk[top]]) L[sk[top]] = j, top --;
            sk[++top] = j;
        }
        for (int j=1; j<=m; ++j) 
            Ans = max(Ans, min(U[i][j], R[j] - L[j] - 1));
    }
    /*memset(R, 0x3f, sizeof(R)); // 悬线法 
    for (int i=1; i<=n; ++i) {
        int last = 0; 
        for (int j=1; j<=m; ++j) {
            if (a[i][j] == '.') L[j] = max(L[j], last + 1);
            else last = j, L[j] = 0;
        }
        last = m + 1;
        for (int j=m; j>=1; --j) {
            if (a[i][j] == '.') R[j] = min(R[j], last - 1);
            else last = j, R[j] = m + 1;
        }
        for (int j=1; j<=m; ++j) 
            Ans = max(Ans, min(U[i][j], R[j] - L[j] + 1));
    }*/
}
bool update(int x,int k) { // 判断能否组成k*k的方阵 
    int L1 = 1, R1 = 0, L2 = 1, R2 = 0;
    for (int j=1; j<=m; ++j) {
        while (L1 <= R1 && j - q1[L1] + 1 > k) L1 ++;
        while (L1 <= R1 && U[x][q1[R1]] >= U[x][j]) R1 --;
        q1[++R1] = j;
        
        while (L2 <= R2 && j - q2[L2] + 1 > k) L2 ++;
        while (L2 <= R2 && D[x][q2[R2]] >= D[x][j]) R2 --;
        q2[++R2] = j;
        
        if (j >= k && U[x][q1[L1]] + D[x][q2[L2]] - 1 >= k) {
            Ans = k;
            return true;
        }
    }
    return false;
}
void Change(int x,int y) {
    a[x][y] = '.';
    for (int i=x; i<=n; ++i) U[i][y] = a[i][y] == '.' ? U[i-1][y] + 1 : 0;
    for (int i=x; i>=1; --i) D[i][y] = a[i][y] == '.' ? D[i+1][y] + 1 : 0;
}
int main() {
    n = read(), m = read(); int Q = read();
    for (int i=1; i<=n; ++i) scanf("%s",a[i]+1);
    for (int i=1; i<=Q; ++i) {
        x[i] = read(),y[i] = read();
        a[x[i]][y[i]] = 'X';
    }
    solve();
    for (int i=Q; i>=1; --i) {
        ans[i] = Ans;
        Change(x[i], y[i]);
        while (update(x[i], Ans + 1));
    }
    for (int i=1; i<=Q; ++i) printf("%d
",ans[i]);
    return 0;
}