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。

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

;
int x
, y
, L
, R
, U

, D

, q1
, q2
, ans
, sk
;
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\n",ans[i]);
return 0;
}