今天做一个微软的校招笔试题 Registration Day ,用优先队列模拟操作的。粘贴来别人的代码,谨记 pq 的用法。另外 memset 包含在 string.h 里。
1 #include <stdio.h> 2 #include <string.h> 3 #include <limits.h> 4 #include <queue> 5 #include <vector> 6 #include <set> 7 #include <map> 8 #include <algorithm> 9 10 using namespace std; 11 12 int N, M, K; 13 struct Student 14 { 15 int id; 16 int arrive_time; 17 int office_num; 18 int finish_time; 19 std::vector<pair<int, int> > register_offices; 20 }; 21 Student students[10005]; 22 struct Event 23 { 24 int student_idx; 25 int office; 26 int begin; 27 int duration; 28 Event(int s, int o, int b, int d): student_idx(s), office(o), begin(b), duration(d){}; 29 }; 30 struct Cmp 31 { 32 bool operator()(const Event &e1, const Event &e2){ 33 if (e1.begin == e2.begin) 34 return students[e1.student_idx].id > students[e2.student_idx].id; 35 return e1.begin > e2.begin; 36 } 37 }; 38 priority_queue<Event, vector<Event>, Cmp> pq; 39 int pre[105]; 40 int pos[10005]; 41 void Solve() { 42 for (int i=0;i<N;++i) { 43 pq.push(Event(i, students[i].register_offices[0].first, students[i].arrive_time+K, students[i].register_offices[0].second)); 44 pos[i] = 1; 45 } 46 memset(pre, -1, sizeof(pre)); 47 while (!pq.empty()) { 48 Event e = pq.top(); 49 pq.pop(); 50 int st = e.student_idx; 51 if (pre[e.office] > e.begin) { 52 e.begin = pre[e.office]; 53 } 54 int finish_time = e.begin + e.duration; 55 if (pos[st] == students[st].office_num) { 56 students[st].finish_time = finish_time; 57 } else { 58 int office = students[st].register_offices[pos[st]].first; 59 int duration = students[st].register_offices[pos[st]].second; 60 pq.push(Event(st, office, finish_time+K, duration)); 61 pos[st]++; 62 } 63 pre[e.office] = finish_time; 64 } 65 } 66 int main() 67 { 68 scanf("%d%d%d", &N, &M, &K); 69 int p, o, w; 70 for (int i=0;i<N;++i) { 71 scanf("%d%d%d", &students[i].id, &students[i].arrive_time, &p); 72 students[i].office_num = p; 73 for (int j=0;j<p;++j) { 74 scanf("%d%d", &o, &w); 75 students[i].register_offices.push_back(make_pair(o, w)); 76 } 77 } 78 Solve(); 79 for (int i=0;i<N;++i) { 80 printf("%d\n", students[i].finish_time); 81 } 82 return 0; 83 }
以下是转载的 functional 模板的优先队列用法,以供方便做题查找
1 #include<iostream> 2 #include<functional> 3 #include<queue> 4 using namespace std; 5 struct node 6 { 7 friend bool operator< (node n1, node n2) 8 { 9 return n1.priority < n2.priority; //"<"为从大到小排列,">"为从小到大排列 10 } 11 int priority; 12 int value; 13 }; 14 int main() 15 { 16 const int len = 5; //也可以写在函数内 17 int i; 18 int a[len] = {3,5,9,6,2}; 19 //示例1 20 priority_queue<int> qi; //普通的优先级队列,按从大到小排序 21 for(i = 0; i < len; i++) 22 qi.push(a[i]); 23 for(i = 0; i < len; i++) 24 { 25 cout<<qi.top()<<" "; 26 qi.pop(); 27 } 28 cout<<endl; 29 //示例2 30 priority_queue<int, vector<int>, greater<int> > qi2; //从小到大的优先级队列,可将greater改为less,即为从大到小 31 for(i = 0; i < len; i++) 32 qi2.push(a[i]); 33 for(i = 0; i < len; i++) 34 { 35 cout<<qi2.top()<<" "; 36 qi2.pop(); 37 } 38 cout<<endl; 39 //示例3 40 priority_queue<node> qn; //必须要重载运算符 41 node b[len]; 42 b[0].priority = 6; b[0].value = 1; 43 b[1].priority = 9; b[1].value = 5; 44 b[2].priority = 2; b[2].value = 3; 45 b[3].priority = 8; b[3].value = 2; 46 b[4].priority = 1; b[4].value = 4; 47 48 for(i = 0; i < len; i++) 49 qn.push(b[i]); 50 cout<<"优先级"<<‘\t‘<<"值"<<endl; 51 for(i = 0; i < len; i++) 52 { 53 cout<<qn.top().priority<<‘\t‘<<qn.top().value<<endl; 54 qn.pop(); 55 } 56 return 0; 57 }
时间: 2024-11-10 10:22:16