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HDU 2883 kebab

题目链接

题意：有一个烧烤机，每次最多能烤 m 块肉，现在有 n 个人来买烤肉，每个人到达时间为 si，离开时间为 ei，点的烤肉数量为 ci，每个烤肉所需烘烤时间为 di，注意一个烤肉可以切成几份来烤

思路：把区间每个点存起来排序后，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，每个时间内只可能有一个任务器做，这样建模就简单了，源点连向汇点，容量为任务需要总时间，区间连向汇点，容量为区间长度，然后每个任务如果包含了某个区间，之间就连边容量无限大，最后判断一下最大流是否等于总任务需要时间即可

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;
}

HDU 2883 kebab（最大流）

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原文地址：http://blog.csdn.net/accelerator_/article/details/40991827