【HNOI2007】紧急疏散

题面

题解

\(\text{HNOI2007}\)真的恐怖

这是集合了所罗门的咒语，胜负一子等神仙题和码农题的一年

所以这道题非常码

二分答案，将门拆点，于是就变成了一个二分图匹配的题目

反正很恶心

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<queue>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x))

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int N(25), INF(0x3f3f3f3f);
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
int n, m; char g[N][N];

namespace DINIC
{
	const int maxn(5e4 + 10), maxm(1e6 + 10);
	struct edge { int next, to, cap; } e[maxm];
	int head[maxn], e_num = -1, S, T, q[maxn], tail, lev[maxn], cur[maxn];
	inline void add_edge(int from, int to, int cap)
	{
		e[++e_num] = (edge) {head[from], to, cap}; head[from] = e_num;
		e[++e_num] = (edge) {head[to], from,  0 }; head[to]   = e_num;
	}

	int bfs()
	{
		clear(lev, 0); q[tail = lev[S] = 1] = S;
		for(RG int i = 1; i <= tail; i++)
		{
			int x = q[i];
			for(RG int j = head[x]; ~j; j = e[j].next)
			{
				int to = e[j].to;
				if(lev[to] || (!e[j].cap)) continue;
				lev[q[++tail] = to] = lev[x] + 1;
			}
		}
		return lev[T];
	}

	int dfs(int x, int f)
	{
		if(x == T || (!f)) return f;
		int ans = 0, cap;
		for(RG int &i = cur[x]; ~i; i = e[i].next)
		{
			int to = e[i].to;
			if(e[i].cap && lev[to] == lev[x] + 1)
			{
				cap = dfs(to, std::min(f - ans, e[i].cap));
				e[i].cap -= cap, e[i ^ 1].cap += cap, ans += cap;
				if(ans == f) break;
			}
		}
		return ans;
	}

	inline int main()
	{
		int ans = 0;
		while(bfs())
		{
			for(RG int i = S; i <= T; i++) cur[i] = head[i];
			ans += dfs(S, INF);
		}
		return ans;
	}
}

std::vector<int> dX, dY, pX, pY;
int dis[N][N][N][N];

void BFS(int x, int y, int d[N][N])
{
	std::queue<int> qx, qy;
	d[x][y] = 0; qx.push(x), qy.push(y);
	while(!qx.empty())
	{
		x = qx.front(); qx.pop();
		y = qy.front(); qy.pop();
		for(RG int k = 0; k < 4; k++)
		{
			int tx = x + dx[k], ty = y + dy[k];
			if(tx < 1 || tx > n || ty < 1 || ty > m) continue;
			if(g[tx][ty] != '.' || d[tx][ty] != -1) continue;
			d[tx][ty] = d[x][y] + 1;
			qx.push(tx), qy.push(ty);
		}
	}
}

bool check(int t)
{
	int d = dX.size(), p = pX.size(); DINIC::e_num = -1;
	clear(DINIC::head, -1); static int idcnt, id_door[N * N][N], id_p[N];
	DINIC::S = idcnt = 1;
	for(RG int i = 1; i <= t; i++)
		for(RG int j = 1; j <= d; j++)
			id_door[i][j] = ++idcnt;
	for(RG int i = 1; i <= t; i++)
		for(RG int j = 1; j <= d; j++)
			DINIC::add_edge(DINIC::S, id_door[i][j], 1);
	for(RG int i = 1; i <= p; i++)
		id_p[i] = ++idcnt;
	DINIC::T = ++idcnt;
	for(RG int i = 1; i <= p; i++)
		DINIC::add_edge(id_p[i], DINIC::T, 1);
	for(RG int i = 0; i < d; i++)
		for(RG int j = 0; j < p; j++)
		{
			int ds = dis[dX[i]][dY[i]][pX[j]][pY[j]];
			if(ds < 0) continue;
			for(RG int k = ds; k <= t; k++)
				DINIC::add_edge(id_door[k][i + 1], id_p[j + 1], 1);
		}
	return DINIC::main() == p;
}

void Solve()
{
	int lim = n * m; clear(dis, -1);
	for(RG int x = 1; x <= n; x++)
		for(RG int y = 1; y <= m; y++)
			if(g[x][y] == 'D') dX.push_back(x),
				dY.push_back(y), BFS(x, y, dis[x][y]);
			else if(g[x][y] == '.') pX.push_back(x), pY.push_back(y);
	int L = -1, R = lim + 1, ans = R;
	while(L <= R)
	{
		int mid = (L + R) >> 1;
		if(check(mid)) R = mid - 1, ans = mid;
		else L = mid + 1;
	}
	if(ans > lim) puts("impossible");
	else printf("%d\n", ans);
}

int main()
{
	n = read(), m = read();
	for(RG int i = 1; i <= n; i++)
		scanf("%s", g[i] + 1);
	Solve();
	return 0;
}