Bzoj4242:水壶

Sol

暴力就是每个建筑物跑一遍 (BFS) 然后最短距离建最小生成树，询问倍增
正解比较巧妙
每个点记录一个 (dis) 表示这个点到最近建筑的距离，(vis) 表示最近的是哪一个
当一个建筑物走到它时，如果有其它的走到了这个点，那么直接两个建筑物连边

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

const int maxn(2005);
const int maxm(4000005);
const int maxp(2e5 + 5);

struct Data{
	int x, y;
};

struct Edge{
	int v, w;
};

int k, n, m, vis[maxm], dis[maxm], qry, deep[maxp];
int fa[maxp], idx, posx[maxm], posy[maxm], st[19][maxp], mx[19][maxp];
int dx[4] = {0, 0, -1, 1}, dy[4] = {-1, 1};
bool wall[maxm];
char s[maxn];
vector <Edge> graph[maxp];
vector <Data> edge[maxm];
queue <int> q;

IL int Find(RG int x){
	return x == fa[x] ? x : fa[x] = Find(fa[x]);
}

IL void Add(RG int u, RG int v, RG int w){
	graph[u].push_back((Edge){v, w});
	graph[v].push_back((Edge){u, w});
}

IL void Dfs(RG int u, RG int ff){
	for(RG int i = 1; i <= 18; ++i){
		st[i][u] = st[i - 1][st[i - 1][u]];
		mx[i][u] = max(mx[i - 1][u], mx[i - 1][st[i - 1][u]]);
	}
	for(RG auto x: graph[u])
		if(x.v != ff){
			mx[0][x.v] = x.w, st[0][x.v] = u;
			deep[x.v] = deep[u] + 1;
			Dfs(x.v, u);
		}
}

# define ID(i, j) ((i - 1) * m + j)

IL int Calc(RG int u, RG int v){
	RG int ret = 0;
	if(deep[u] < deep[v]) swap(u, v);
	for(RG int j = 18; ~j; --j)
		if(deep[st[j][u]] >= deep[v]){
			ret = max(ret, mx[j][u]);
			u = st[j][u];
		}
	if(u == v) return ret;
	for(RG int j = 18; ~j; --j)
		if(st[j][u] != st[j][v]){
			ret = max(ret, max(mx[j][u], mx[j][v]));
			u = st[j][u], v = st[j][v];
		}
	return max(ret, max(mx[0][u], mx[0][v]));
}

int main(){
	n = Input(), m = Input(), k = Input(), qry = Input();
	for(RG int i = 1; i <= n; ++i){
		scanf(" %s", s + 1);
		for(RG int j = 1; j <= m; ++j){
			++idx, posx[idx] = i, posy[idx] = j;
			wall[idx] = s[j] == '#';
		}
	}
	for(RG int i = 1; i <= k; ++i){
		RG int px = Input(), py = Input(), id = ID(px, py);
		vis[id] = fa[i] = i, dis[id] = 0, q.push(id);
	}
	while(!q.empty()){
		RG int u = q.front(); q.pop();
		for(RG int i = 0; i < 4; ++i){
			RG int xx = posx[u] + dx[i], yy = posy[u] + dy[i];
			if(xx < 1 || yy < 1 || xx > n || yy > m) continue;
			RG int id = ID(xx, yy);
			if(wall[id]) continue;
			if(!vis[id]){
				vis[id] = vis[u], dis[id] = dis[u] + 1;
				q.push(id);
			}
			else edge[dis[id] + dis[u]].push_back((Data){vis[u], vis[id]});
		}
	}
	for(RG int i = 0; i <= n * m; ++i)
		for(RG auto v: edge[i]){
			RG int fx = Find(v.x), fy = Find(v.y);
			if(fx != fy) fa[fx] = fy, Add(v.x, v.y, i);
		}
	for(RG int i = 1; i <= k; ++i) if(!st[0][i]) Dfs(i, 0);
	while(qry--){
		RG int u = Input(), v = Input();
		if(Find(u) != Find(v)) puts("-1");
		else printf("%d
", Calc(u, v));
	}
	return 0;
}