Type \(1\) 是简单的。直接输出空格个数即可。
Type \(2\) 也是简单的。显然要堵住不在起点和出口最短路上的格子，答案为空格个数减去起点到任一出口的最短路。
考虑 Type \(3\)。容易发现答案为空格个数减去起点到任两个出口的最短路（公共部分只算一次）。考虑从起点开始出发，一定最多存在一个格子，使得从起点走到它后，两条路径分道扬镳，不再相交。那我们可以枚举这个格子，求出它到最近的两个不同出口的距离（就是最小和次小）。可以简单 bfs 求出。
总时间复杂度 \(O(nm)\)。
code
// Problem: F. Vova Escapes the Matrix
// Contest: Codeforces - Codeforces Round 910 (Div. 2)
// URL: https://codeforces.com/contest/1898/problem/F
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<int, int> pii;
const int maxn = 1010;
const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
int n, m, f[maxn][maxn][2], g[maxn][maxn];
bool vis[maxn][maxn][2], mk[maxn][maxn];
pii h[maxn][maxn][2];
char s[maxn][maxn];
struct node {
int x, y, o;
node(int a = 0, int b = 0, int c = 0) : x(a), y(b), o(c) {}
};
void solve() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%s", s[i] + 1);
}
int sx = -1, sy = -1;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
f[i][j][0] = f[i][j][1] = g[i][j] = 1e8;
vis[i][j][0] = vis[i][j][1] = mk[i][j] = 0;
h[i][j][0] = h[i][j][1] = mkp(0, 0);
if (s[i][j] == 'V') {
sx = i;
sy = j;
}
}
}
queue<node> q;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if ((i == 1 || i == n || j == 1 || j == m) && s[i][j] != '#') {
f[i][j][0] = 0;
vis[i][j][0] = 1;
h[i][j][0] = mkp(i, j);
q.emplace(i, j, 0);
}
}
}
while (q.size()) {
int x = q.front().x, y = q.front().y, o = q.front().o;
q.pop();
for (int i = 0; i < 4; ++i) {
int nx = x + dx[i], ny = y + dy[i];
if (nx < 1 || nx > n || ny < 1 || ny > m || s[nx][ny] == '#') {
continue;
}
if (!h[nx][ny][0].fst) {
h[nx][ny][0] = h[x][y][o];
f[nx][ny][0] = f[x][y][o] + 1;
q.emplace(nx, ny, 0);
} else if (h[nx][ny][0] != h[x][y][o] && !h[nx][ny][1].fst) {
h[nx][ny][1] = h[x][y][o];
f[nx][ny][1] = f[x][y][o] + 1;
q.emplace(nx, ny, 1);
}
}
}
static pii Q[maxn * maxn];
int hd = 1, tl = 0;
Q[++tl] = mkp(sx, sy);
mk[sx][sy] = 1;
g[sx][sy] = 0;
while (hd <= tl) {
int x = Q[hd].fst, y = Q[hd].scd;
++hd;
for (int i = 0; i < 4; ++i) {
int nx = x + dx[i], ny = y + dy[i];
if (nx < 1 || nx > n || ny < 1 || ny > m || s[nx][ny] == '#') {
continue;
}
if (!mk[nx][ny]) {
mk[nx][ny] = 1;
g[nx][ny] = g[x][y] + 1;
Q[++tl] = mkp(nx, ny);
}
}
}
int cnt = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if (i == 1 || i == n || j == 1 || j == m) {
cnt += mk[i][j];
}
}
}
int t = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
t += (s[i][j] == '.');
}
}
if (cnt == 0) {
printf("%d\n", t);
return;
}
if (cnt == 1) {
int mn = 1e8;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if (i == 1 || i == n || j == 1 || j == m) {
mn = min(mn, g[i][j]);
}
}
}
printf("%d\n", t - mn);
return;
}
int ans = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
ans = max(ans, t - g[i][j] - f[i][j][0] - f[i][j][1]);
}
}
printf("%d\n", ans);
}
int main() {
int T = 1;
scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}