POJ 3762 The Bonus Salary!（最小K覆盖）

POJ 3762 The Bonus Salary!

题目链接

题意：给定一些任务，每个任务有一个时间，有k天，一个时间只能执行一个任务，每个任务有一个价值，问怎么安排能得到最多价值

思路：典型的区间k覆盖问题

代码：

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>#include <map>using namespace std;const int MAXNODE = 4505;const int MAXEDGE = 100005;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 = 2005;map<int, int> hash;int n, k, hn;int get(int x) {if (!hash.count(x)) hash[x] = hn++;return hash[x];}struct Task {int l, r, w;} task[N];int main() {while (~scanf("%d%d", &n, &k)) {hash.clear();hn = 1;int h1, m1, s1, h2, m2, s2, w;for (int i = 0; i < n; i++) {scanf("%d:%d:%d %d:%d:%d %d", &h1, &m1, &s1, &h2, &m2, &s2, &w);int t1 = h1 * 3600 + m1 * 60 + s1;int t2 = h2 * 3600 + m2 * 60 + s2;task[i].l = get(t1); task[i].r = get(t2); task[i].w = w;}gao.init(hn + 1);for (int i = 0; i < n; i++)gao.add_Edge(task[i].l, task[i].r, 1, -task[i].w);int pre = 0;for (map<int, int>::iterator it = hash.begin(); it != hash.end(); it++) {gao.add_Edge(pre, it->second, k, 0);pre = it->second;}gao.add_Edge(pre, hn, k, 0);printf("%d\n", -gao.Mincost(0, hn));}return 0;}