HDU 2732 Leapin‘ Lizards

题目链接

题意：有一些蜥蜴在一个迷宫里面，有一个跳跃力表示能跳到多远的柱子，然后每根柱子最多被跳一定次数。求这些蜥蜴还有多少是不管怎样都逃不出来的。

思路：把柱子拆点建图跑最大流就可以。还是挺明显的

代码：

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

const int MAXNODE = 805;
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;

	void init(int n) {
		this->n = n;
		memset(first, -1, sizeof(first));
		m = 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;
		Type flow = 0;
		while (bfs()) {
			for (int i = 0; i < n; i++)
				cur[i] = first[i];
			flow += dfs(s, INF);
		}
		return flow;
	}

	void MinCut() {
		cut.clear();
		for (int i = 0; i < m; i += 2) {
			if (vis[edges[i].u] && !vis[edges[i].v])
				cut.push_back(i);
		}
	}
} gao;

const int N = 25;

int T, n, m;
double d;
char str[N];

int main() {
	int cas = 0;
	scanf("%d", &T);
	while (T--) {
		scanf("%d%lf", &n, &d);
		gao.init(n * 20 * 2 + 2);
		int s = n * 20 * 2, t = n * 20 * 2 + 1;
		for (int i = 0; i < n; i++) {
			scanf("%s", str);
			m = strlen(str);
			for (int j = 0; j < m; j++) {
				if (str[j] != ‘0‘)
					gao.add_Edge(i * m + j, i * m + j + n * m, str[j] - ‘0‘);
				if (i - d < 0 || i + d >= n || j - d < 0 || j + d >= m)
					gao.add_Edge(i * m + j + n * m, t, INF);
			}
		}
		int tot = 0;
		for (int i = 0; i < n; i++) {
			scanf("%s", str);
			m = strlen(str);
			for (int j = 0; j < m; j++) {
				if (str[j] == ‘L‘) {
					tot++;
					gao.add_Edge(s, i * m + j, 1);
				}
			}
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				for (int x = 0; x < n; x++) {
					for (int y = 0; y < m; y++) {
						int dx = i - x;
						int dy = j - y;
						double dis = sqrt(dx * dx * 1.0 + dy * dy);
						if (dis > d) continue;
						gao.add_Edge(i * m + j + n * m, x * m + y, INF);
					}
				}
			}
		}
		printf("Case #%d: ", ++cas);
		int ans = tot - gao.Maxflow(s, t);
		if (ans == 0) printf("no ");
		else printf("%d ", ans);
		if (ans <= 1) printf("lizard was ");
		else printf("lizards were ");
		printf("left behind.\n");
	}
	return 0;
}