HDU 2883 kebab

题目链接

题意：有一个烧烤机，每次最多能烤 m 块肉。如今有 n 个人来买烤肉，每一个人到达时间为 si。离开时间为 ei，点的烤肉数量为 ci，每一个烤肉所需烘烤时间为 di。注意一个烤肉能够切成几份来烤

思路：把区间每一个点存起来排序后。得到最多2 * n - 1个区间，这些就表示几个互相不干扰的时间，每一个时间内仅仅可能有一个任务器做。这样建模就简单了。源点连向汇点，容量为任务须要总时间，区间连向汇点，容量为区间长度。然后每一个任务假设包括了某个区间，之间就连边容量无限大。最后推断一下最大流是否等于总任务须要时间就可以

代码：

#include <cstdio>

#include <cstring>

#include <queue>

#include <algorithm>

using namespace std;

const int MAXNODE = 1005;

const int MAXEDGE = 200005;

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 = 405;

int n, m;

struct Man {

	int l, r;

} man[N];

int p[N], pn;

int s, nu, e, t;

int main() {

	while (~scanf("%d%d", &n, &m)) {

		gao.init(3 * n + 2);

		pn = 0;

		int sum = 0;

		for (int i = 1; i <= n; i++) {

			scanf("%d%d%d%d", &s, &nu, &e, &t);

			man[i].l = s; man[i].r = e;

			p[pn++] = s; p[pn++] = e;

			sum += nu * t;

			gao.add_Edge(0, i, nu * t);

		}

		sort(p, p + pn);

		for (int i = 1; i < pn; i++)

			gao.add_Edge(n + i, 3 * n + 1, (p[i] - p[i - 1]) * m);

		for (int i = 1; i <= n; i++) {

			for (int j = 1; j < pn; j++) {

				if (p[j - 1] > man[i].r) break;

				if (man[i].l <= p[j - 1] && man[i].r >= p[j])

					gao.add_Edge(i, j + n, INF);

			}

		}

		printf("%s\n", gao.Maxflow(0, 3 * n + 1) ==  sum ? "Yes" : "No");

	}

	return 0;

}