UVA 11380 - Down Went The Titanic

题目链接

题意：给定一个图，上面有薄冰‘.‘或‘*‘，厚冰‘@‘,木块‘#‘，一开始人都在‘*‘上，薄冰只能走一次就会沉掉，厚冰次数不限，如果人走到木块上就获救了，但是一个木块的容量只有p，求最多能有多少人获救

思路：最大流，由于点有次数限制，所以可以进行拆点，然后建图每个4和四个方向建边，源点和‘*‘建边，‘#‘和汇点建边容量为p，然后跑一下最大流即可

代码：

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

const int MAXNODE = 2005;
const int MAXEDGE = 100005;

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 = 35;
const int d[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};

int x, y, p;
char str[N];

int main() {
	while (~scanf("%d%d%d", &x, &y, &p)) {
		int tot = x * y;
		gao.init(2 * tot + 2);
		for (int i = 0; i < x; i++) {
			scanf("%s", str);
			for (int j = 0; j < y; j++) {
				int u = i * y + j + 1;
				if (str[j] == '*' || str[j] == '.') {
					gao.add_Edge(u, u + tot, 1);
					if (str[j] == '*')
						gao.add_Edge(0, u, 1);
				}
				if (str[j] == '#') {
					gao.add_Edge(u, u + tot, INF);
					gao.add_Edge(u + tot, 2 * tot + 1, p);
				}
				if (str[j] == '@') gao.add_Edge(u, u + tot, INF);
				for (int k = 0; k < 4; k++) {
					int xx = i + d[k][0];
					int yy = j + d[k][1];
					if (xx < 0 || xx >= x || yy < 0 || yy >= y) continue;
					int v = xx * y + yy + 1;
					gao.add_Edge(u + tot, v, INF);
				}
			}
		}
		printf("%d\n", gao.Maxflow(0, 2 * tot + 1));
	}
	return 0;
}