UVA 1317 - Concert Hall Scheduling（网络流）

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UVA 1317 - Concert Hall Scheduling

题目链接

题意：现在有两个音乐厅，有一些人要租用，每次租一个区间的时间，给w钱，要求一个租的方案使得总收入最大，问总收入

思路：区间k覆盖问题，一个左闭右开区间可以建一条边，容量为1，代价为-w（因为要求最大），然后区间每个[i, i + 1]建一条边，容量2，代价0，然后跑一下费用流即可

代码：

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;const int MAXNODE = 400;const int MAXEDGE = 5005 * 4;typedef int Type;const Type INF = 0x3f3f3f3f;struct Edge {int u, v;Type cap, flow, cost;Edge() {}Edge(int u, int v, Type cap, Type flow, Type cost) {this->u = u;this->v = v;this->cap = cap;this->flow = flow;this->cost = cost;}};struct MCFC {int n, m, s, t;Edge edges[MAXEDGE];int first[MAXNODE];int next[MAXEDGE];int inq[MAXNODE];Type d[MAXNODE];int p[MAXNODE];Type a[MAXNODE];void init(int n) {this->n = n;memset(first, -1, sizeof(first));m = 0;}void add_Edge(int u, int v, Type cap, Type cost) {edges[m] = Edge(u, v, cap, 0, cost);next[m] = first[u];first[u] = m++;edges[m] = Edge(v, u, 0, 0, -cost);next[m] = first[v];first[v] = m++;}bool bellmanford(int s, int t, Type &flow, Type &cost) {for (int i = 0; i < n; i++) d[i] = INF;memset(inq, false, sizeof(inq));d[s] = 0; inq[s] = true; p[s] = s; a[s] = INF;queue<int> Q;Q.push(s);while (!Q.empty()) {int u = Q.front(); Q.pop();inq[u] = false;for (int i = first[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (e.cap > e.flow && d[e.v] > d[u] + e.cost) {d[e.v] = d[u] + e.cost;p[e.v] = i;a[e.v] = min(a[u], e.cap - e.flow);if (!inq[e.v]) {Q.push(e.v); inq[e.v] = true;}}}}if (d[t] == INF) return false;flow += a[t];cost += d[t] * a[t];int u = t;while (u != s) {edges[p[u]].flow += a[t];edges[p[u]^1].flow -= a[t];u = edges[p[u]].u;}return true;}Type Mincost(int s, int t) {Type flow = 0, cost = 0;while (bellmanford(s, t, flow, cost));return cost;}} gao;const int N = 1005;int n;int main() {while (~scanf("%d", &n) && n) {gao.init(367);for (int i = 0; i < 366; i++)gao.add_Edge(i, i + 1, 2, 0);int u, v, w;while (n--) {scanf("%d%d%d", &u, &v, &w);gao.add_Edge(u, v + 1, 1, -w);}printf("%d\n", -gao.Mincost(0, 366));}return 0;}

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