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;
}