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=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;

 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=; i<=n; ++i)
         for (int j=; j<=m; ++j)
             U[i][j] = a[i][j] == '.' ? U[i-][j] +  : ;
     for (int i=n; i>=; --i)
         for (int j=; j<=m; ++j)
             D[i][j] = a[i][j] == '.' ? D[i+][j] +  : ;
     for (int i=; i<=n; ++i) { // 单调栈
         U[i][] = U[i][m+] = -;
         for (int top=,j=; j<=m+; ++j) {
             while (top >  && U[i][j] < U[i][sk[top]]) R[sk[top]] = j, top--;
             sk[++top] = j;
         }
         for (int top=,j=m; j>=; --j) {
             while (top >  && U[i][j] < U[i][sk[top]]) L[sk[top]] = j, top --;
             sk[++top] = j;
         }
         for (int j=; j<=m; ++j)
             Ans = max(Ans, min(U[i][j], R[j] - L[j] - ));
     }
     /*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 = , R1 = , L2 = , R2 = ;
     for (int j=; j<=m; ++j) {
         while (L1 <= R1 && j - q1[L1] +  > k) L1 ++;
         while (L1 <= R1 && U[x][q1[R1]] >= U[x][j]) R1 --;
         q1[++R1] = j;

         while (L2 <= R2 && j - q2[L2] +  > 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]] -  >= 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-][y] +  : ;
     for (int i=x; i>=; --i) D[i][y] = a[i][y] == '.' ? D[i+][y] +  : ;
 }
 int main() {
     n = read(), m = read(); int Q = read();
     for (int i=; i<=n; ++i) scanf("%s",a[i]+);
     for (int i=; i<=Q; ++i) {
         x[i] = read(),y[i] = read();
         a[x[i]][y[i]] = 'X';
     }
     solve();
     for (int i=Q; i>=; --i) {
         ans[i] = Ans;
         Change(x[i], y[i]);
         while (update(x[i], Ans + ));
     }
     for (int i=; i<=Q; ++i) printf("%d\n",ans[i]);
     return ;
 }