Codeforces 679C Bear and Square Grid(滑动窗口)

题意：

给定一个网格图，其中有的有的是可以走的点，有的是障碍，定义连通块和无向图中连通块是一样的。如果把一个k*k大小的区域内的点都变成联通块，那么这个网格图中最大的连通块是多大

解法：

这个如果用暴力枚举的话，是O(N4)的，但是我们可以用滑动窗口来优化，我们首先把每一个连通块的大小求出来，然后枚举y，先把窗口内的点全部变为联通，并且从他所在的连通块分离出去，那么我们只需要检测当前连通块的上下左右四条边所连接的连通块来更新答案即可，因为一个连通块可能被统计很多次，所以我们要给每一个连通块打一个时间戳，来标记是否被当前情况添加过。具体可以看代码。

////  Created by  CQU_CST_WuErli//  Copyright (c) 2016 CQU_CST_WuErli. All rights reserved.////#pragma comment(linker, "/STACK:102400000,102400000")#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <string>#include <vector>#include <map>#include <queue>#include <stack>#include <set>#include <algorithm>#include <sstream>#define CLR(x) memset(x,0,sizeof(x))#define OFF(x) memset(x,-1,sizeof(x))#define MEM(x,a) memset((x),(a),sizeof(x))#define BUG cout << "I am here" << endl#define lookln(x) cout << #x << "=" << x << endl#define SI(a) scanf("%d", &a)#define SII(a,b) scanf("%d%d", &a, &b)#define SIII(a,b,c) scanf("%d%d%d", &a, &b, &c)const int INF_INT=0x3f3f3f3f;const long long INF_LL=0x7f7f7f7f;const int MOD=1e9+7;const double eps=1e-10;const double pi=acos(-1);typedef long long  ll;using namespace std;int n, k;char mp[555][555];int vis[555][555];int size[555 * 555];int tid[555 * 555];int cnt;int dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};int sum;void dfs(int x, int y) {    vis[x][y] = cnt;    sum++;    for (int i = 0; i < 4; i++) {        int tx = x + dir[i][0], ty = y + dir[i][1];        if (tx < 1 || ty < 1 || tx > n || ty > n || mp[tx][ty] == 'X' || vis[tx][ty]) continue;        dfs(tx, ty);    }}void add(int x, int y, int ti, int& ans) {    if (x < 1 || x > n || y < 1 || y > n || mp[x][y] == 'X') return;    if (tid[vis[x][y]] != ti) {        ans += size[vis[x][y]];        tid[vis[x][y]] = ti;    }}int main(int argc, char const *argv[]) {#ifdef LOCAL    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);    // freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);#endif    while(SII(n, k) ==2) {        for (int i = 1; i <= n; i++)            scanf("%s", mp[i] + 1);        CLR(vis);        CLR(size);        CLR(tid);        int ti = 1;        cnt = 0;        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= n; j++) if (mp[i][j] == '.' && !vis[i][j]) {                cnt++;                sum = 0;                dfs(i, j);                size[cnt] = sum;            }        }        int ans = 0;        for (int y = 1; y + k - 1<= n; y++) {            for (int x = 1; x <= k; x++) {                for (int i = 0; i < k; i++) {                    size[vis[x][y + i]]--;                }            }            for (int x = 1; x + k - 1<= n; x++) {                int tmp = k * k;                for (int i = x; i < x + k; i++) {                    add(i, y - 1, ti, tmp);                    add(i, y + k, ti, tmp);                }                for (int i = y; i < y + k; i++) {                    add(x - 1, i, ti, tmp);                    add(x + k, i, ti, tmp);                }                ti++;                ans = max(tmp, ans);                if (x + k <= n) {                    for (int i = y; i < y + k; i++) {                        size[vis[x][i]]++;                        size[vis[x + k][i]]--;                    }                }            }            for (int i = n - k + 1; i <= n; i++) {                for (int j = y; j < y + k; j++) {                    size[vis[i][j]]++;                }            }        }        cout << ans << endl;    }    return 0;}