HDU 1733 Escape

题目链接

题意：给定一个图，#是墙，@是出口，.可以行走，X是人，每个时间每个格子只能站一个人，问最少需要多少时间能让人全部撤离（从出口出去）

思路：网络流，把每个结点每秒当成一个结点，这样枚举时间，每多一秒就在原来的网络上直接加一层继续增广即可，注意考虑方向的时候，要考虑上原地不动

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 100005;
const int MAXEDGE = 500005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v;
	Type cap, flow;
	Edge() {}
	Edge(int u, int v, Type cap, Type flow) {
		this->u = u;
		this->v = v;
		this->cap = cap;
		this->flow = flow;
	}
};

struct Dinic {
	int n, m, s, t;
	Edge edges[MAXEDGE];
	int first[MAXNODE];
	int next[MAXEDGE];
	bool vis[MAXNODE];
	Type d[MAXNODE];
	int cur[MAXNODE];
	vector<int> cut;
	Type flow;

	void init(int n) {
		this->n = n;
		memset(first, -1, sizeof(first));
		m = 0;
		flow = 0;
	}
	void add_Edge(int u, int v, Type cap) {
		edges[m] = Edge(u, v, cap, 0);
		next[m] = first[u];
		first[u] = m++;
		edges[m] = Edge(v, u, 0, 0);
		next[m] = first[v];
		first[v] = m++;
	}

	bool bfs() {
		memset(vis, false, sizeof(vis));
		queue<int> Q;
		Q.push(s);
		d[s] = 0;
		vis[s] = true;
		while (!Q.empty()) {
			int u = Q.front(); Q.pop();
			for (int i = first[u]; i != -1; i = next[i]) {
				Edge& e = edges[i];
				if (!vis[e.v] && e.cap > e.flow) {
					vis[e.v] = true;
					d[e.v] = d[u] + 1;
					Q.push(e.v);
				}
			}
		}
		return vis[t];
	}

	Type dfs(int u, Type a) {
		if (u == t || a == 0) return a;
		Type flow = 0, f;
		for (int &i = cur[u]; i != -1; i = next[i]) {
			Edge& e = edges[i];
			if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
				e.flow += f;
				edges[i^1].flow -= f;
				flow += f;
				a -= f;
				if (a == 0) break;
			}
		}
		return flow;
	}

	Type Maxflow(int s, int t) {
		this->s = s; this->t = t;
		while (bfs()) {
			for (int i = 0; i < n; i++)
				cur[i] = first[i];
			flow += dfs(s, INF);
		}
		return flow;
	}
} gao;

#define MP(a,b) make_pair(a,b)
const int N = 20;
const int d[5][2] = {0, 1, 0, -1, 1, 0, -1, 0, 0, 0};

int n, m;
char str[N][N];
typedef pair<int, int> pii;
bool vis[N][N];

bool bfs(int sx, int sy) {
	queue<pii> Q;
	memset(vis, false, sizeof(vis));
	vis[sx][sy] = true;
	Q.push(MP(sx, sy));
	while (!Q.empty()) {
		pii u = Q.front();
		if (str[u.first][u.second] == '@') return true;
		Q.pop();
		for (int i = 0; i < 4; i++) {
			int x = u.first + d[i][0];
			int y = u.second + d[i][1];
			if (x < 0 || x >= n || y < 0 || y >= m || vis[x][y] || str[x][y] == '#') continue;
			vis[x][y] = true;
			Q.push(MP(x, y));
		}
	}
	return false;
}

bool judge() {
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			if (str[i][j] == 'X')
				if (!bfs(i, j)) return false;
		}
	}
	return true;
}

int main() {
	while (~scanf("%d%d", &n, &m)) {
		int tot = 0;
		int s = n * m * 2 * 100, t = n * m * 2 * 100 + 1;
		gao.init(n * m * 2 * 100 + 2);
		for (int i = 0; i < n; i++) {
			scanf("%s", str[i]);
			for (int j = 0; j < m; j++)
				if (str[i][j] == 'X') {
					gao.add_Edge(s, i * m + j, 1);
					tot++;
				}
		}
		if (!judge()) printf("-1\n");
		else {
			for (int ti = 0; ti <= 100; ti++) {
				for (int i = 0; i < n; i++) {
					for (int j = 0; j < m; j++) {
						if (str[i][j] == '#') continue;
						int uin = ti * n * m * 2 + i * m + j;
						int uout = uin + n * m;
						gao.add_Edge(uin, uout, 1);
						if (str[i][j] == '@') gao.add_Edge(uout, t, 1);
						for (int k = 0; k < 5; k++) {
							int x = i + d[k][0];
							int y = j + d[k][1];
							if (x < 0 || x >= n || y < 0 || y >= m || str[x][y] == '#') continue;
							int vin = (ti + 1) * n * m * 2 + x * m + y;
							int vout = vin + n * m;
							gao.add_Edge(uout, vin, 1);
						}
					}
				}
				if (gao.Maxflow(s, t) == tot) {
					printf("%d\n", ti);
					break;
				}
			}
		}
	}
	return 0;
}