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POJ 1698 Alice‘s Chance

题目链接

题意：拍n部电影，每部电影要在前w星期完成，并且一周只有一些天是可以拍的，每部电影有个需要的总时间，问是否能拍完电影

思路：源点向每部电影连边，容量为d，然后每部电影对应能拍的那天连边，由于每天容量限制是1，所以进行拆点，然后连向汇点即可

代码：

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

const int MAXNODE = 1005;
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;

int t, n, day[10], d, w, vis[400];

int main() {
	scanf("%d", &t);
	while (t--) {
		memset(vis, 0, sizeof(vis));
		gao.init(1000);
		scanf("%d", &n);
		int sum = 0;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= 7; j++)
				scanf("%d", &day[j]);
			scanf("%d%d", &d, &w);
			sum += d;
			gao.add_Edge(0, i, d);
			for (int j = 0; j < w; j++) {
				for (int k = 1; k <= 7; k++) {
					if (day[k]) {
						gao.add_Edge(i, 20 + j * 7 + k, INF);
						vis[20 + j * 7 + k] = 1;
					}
				}
			}
		}
		for (int i = 21; i <= 370; i++) {
			if (!vis[i]) continue;
			gao.add_Edge(i, i + 350, 1);
			gao.add_Edge(i + 350, 1000, INF);
		}
		printf("%s\n", gao.Maxflow(0, 1000) == sum ? "Yes" : "No");
	}
	return 0;
}

POJ 1698 Alice's Chance（最大流+拆点）

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原文地址：http://blog.csdn.net/accelerator_/article/details/40818253