Paint the Grid Reloaded (将联通块缩点+BFS)

传送门

题意

能够上下左右相通的并且颜色相同的块为一个整体，就是所谓的联通块。然后每次对联通块进行反转颜色操作，最后要求最小的步数使得图中所有的颜色相同。

思路

• 将所有能够相通的相同颜色进行缩点，形成联通块
• 然后能够上下左右连通的联通块进行建边
• 最后枚举所有的联通块，进行 $bfs$ 标记操作，将所有的答案取最小值，就是最终答案

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
template <typename T>
inline void rd(T& x)
{
    
    
	ll tmp = 1; char c = getchar(); x = 0;
	while (c > '9' || c < '0') {
    
     if (c == '-')tmp = -1; c = getchar(); }
	while (c >= '0' && c <= '9') {
    
     x = x * 10 + c - '0'; c = getchar(); }
	x *= tmp;
}
const int inf = 0x3f3f3f3f;
const int mod = 7;
const int N = 2e4 + 10;
const int M = 1e7 + 10;
const double eps = 1e-8;
int n, m;
char mp[55][55];
bool vis[55][55];
int num[55][55];
int dir[][2] = {
    
     1,0,-1,0,0,1,0,-1 }, cnt;
void bfs(int x,int y) {
    
    
	queue<pii>que;
	que.push({
    
     x,y });
	stack<pii>stk;
	while (!que.empty()) {
    
    
		pii p = que.front(); que.pop();
		stk.push(p);
		if (vis[p.first][p.second]) continue;
		vis[p.first][p.second] = 1;
		for (int i = 0; i < 4; ++i) {
    
    
			int xx = p.first + dir[i][0], yy = p.second + dir[i][1];
			if (xx <= 0 || xx > n || yy <= 0 || yy > m || vis[xx][yy] || mp[xx][yy] != mp[p.first][p.second]) continue;
			que.push({
    
     xx,yy });
		}
	}
	++cnt;
	while (!stk.empty()) {
    
    
		pii p = stk.top(); stk.pop();
		num[p.first][p.second] = cnt;
	}
}
int head[N], cntE;
struct edge {
    
    
	int next, to;
}e[M];
void add(int u, int v) {
    
    
	e[cntE].to = v;
	e[cntE].next = head[u];
	head[u] = cntE++;
}
bool mark[N];
int bfs(int x) {
    
    
	queue<pii>que;
	que.push({
    
     x,1 });
	int ans = 0;
	while (!que.empty()) {
    
    
		pii p = que.front(); que.pop();
		int u = p.first;
		if (mark[u]) continue;
		mark[u] = true;
		ans = max(ans, p.second);
		for (int i = head[u]; ~i; i = e[i].next) {
    
    
			int v = e[i].to;
			if (mark[v]) continue;
			que.push({
    
     v,p.second + 1 });
		}
	}
	return ans-1;
}
int solve() {
    
    
	memset(vis, false, sizeof vis); cnt = 0;
	for (int i = 1; i <= n; ++i) {
    
    
		for (int j = 1; j <= m; ++j) {
    
    
			if (vis[i][j]) continue;
			bfs(i, j);
		}
	}
	memset(head, -1, sizeof head); cntE = 0;
	for (int i = 1; i <= n; ++i) {
    
    
		for (int j = 1; j <= m; ++j) {
    
    
			for (int k = 0; k < 4; ++k) {
    
    
				int x = i + dir[k][0], y = j + dir[k][1];
				if (x <= 0 || x > n || y <= 0 || y > m) continue;
				if (num[x][y] != num[i][j]) add(num[i][j], num[x][y]);
			}
		}
	}
	int minn = inf;
	for (int i = 1; i <= cnt; ++i) {
    
    
		for (int j = 1; j <= cnt; ++j) mark[j] = false;
		minn = min(minn, bfs(i));
	}
	return minn;
}
int main() {
    
    
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
//	FILE* _INPUT = freopen("input.txt", "r", stdin);
//	FILE* _OUTPUT = freopen("output.txt", "w", stdout);
	int T; rd(T);
	while (T--) {
    
    
		rd(n), rd(m);
		for (int i = 1; i <= n; ++i) {
    
    
			for (int j = 1; j <= m; ++j) {
    
    
				scanf(" %c", &mp[i][j]);
			}
		}
		printf("%d\n", solve());
	}
	return 0;
}