Codeforces 679C Bear and Square Grid

Bear and Square Grid

枚举k * k 的位置， 然后接上它周围白色连通块的数量， 再统计完全在k * k范围里的连通块， 这个只要某个连通块全部的方格

在k * k里面就好， 并且k * k是一行一行移的， 所以可以优化到n ^ 3。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 500 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, k, idx, now, id[N][N], sum[N][N];
bool vis[N * N];
char Map[N][N];

int fa[N * N], cnt[N * N], num[N * N];

int getRoot(int x) {
    return fa[x] == x ? x : fa[x] = getRoot(fa[x]);
}

void change(int x, int op) {
    x = getRoot(x);
    if(op == 1) {
        num[x]--;
        if(!num[x]) now += cnt[x];
    } else {
        if(!num[x]) now -= cnt[x];
        num[x]++;
    }
}
int main() {
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n * n; i++) fa[i] = i;
    for(int i = 1; i <= n; i++)
        scanf("%s", Map[i] + 1);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            if(Map[i][j] == '.') id[i][j] = ++idx, cnt[idx] = 1;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            if(Map[i][j] != '.') continue;
            if(Map[i - 1][j] == '.') {
                int x = getRoot(id[i][j]);
                int y = getRoot(id[i - 1][j]);
                if(x != y) fa[y] = x, cnt[x] += cnt[y];
            }
            if(Map[i][j - 1] == '.') {
                int x = getRoot(id[i][j]);
                int y = getRoot(id[i][j - 1]);
                if(x != y) fa[y] = x, cnt[x] += cnt[y];
            }
        }
    }
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + (Map[i][j] == 'X');
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            if(Map[i][j] == '.') num[getRoot(id[i][j])]++;
    
    int ans = 0;
    queue<int> que;

    for(int i = 1; i + k - 1 <= n; i++) {
        for(int j = 1; j + k - 1 <= n; j++) {
            if(j == 1) {
                for(int u = i; u < i + k; u++)
                    for(int v = j; v < j + k; v++)
                        change(id[u][v], 1);
            }
            int ret = 0;
            if(i - 1 > 0) {
                for(int z = j; z <= j + k - 1; z++) {
                    if(Map[i - 1][z] != '.') continue;
                    int x = getRoot(id[i - 1][z]);
                    if(!vis[x]) {
                        vis[x] = true;
                        que.push(x);
                        ret += cnt[x];
                    }
                }
            }
            if(i + k <= n) {
                for(int z = j; z <= j + k - 1; z++) {
                    if(Map[i + k][z] != '.') continue;
                    int x = getRoot(id[i + k][z]);
                    if(!vis[x]) {
                        vis[x] = true;
                        que.push(x);
                        ret += cnt[x];
                    }
                }
            }

            if(j - 1 > 0) {
                for(int z = i; z <= i + k - 1; z++) {
                    if(Map[z][j - 1] != '.') continue;
                    int x = getRoot(id[z][j - 1]);
                    if(!vis[x]) {
                        vis[x] = true;
                        que.push(x);
                        ret += cnt[x];
                    }
                }
            }
            if(j + k <= n) {
                for(int z = i; z <= i + k - 1; z++) {
                    if(Map[z][j + k] != '.') continue;
                    int x = getRoot(id[z][j + k]);
                    if(!vis[x]) {
                        vis[x] = true;
                        que.push(x);
                        ret += cnt[x];
                    }
                }
            }
            while(!que.empty()) {
                vis[que.front()] = false;
                que.pop();
            }
            ans = max(ans, now + ret + sum[i + k - 1][j + k - 1] - sum[i - 1][j + k - 1] - sum[i + k - 1][j - 1] + sum[i - 1][j - 1]);
            if(j + k <= n) {
                for(int z = i; z < i + k; z++) change(id[z][j], -1);
                for(int z = i; z < i + k; z++) change(id[z][j + k], 1);
            } else {
                for(int u = i; u < i + k; u++)
                    for(int v = j; v < j + k; v++)
                        change(id[u][v], -1);
            }
        }
    }
    printf("%d\n", ans);
    return 0;
 }

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