2017微软秋季校园招聘在线编程笔试-#1401 : Registration Day

来源：互联网发布：天猫旗舰店的流量数据编辑：程序博客网时间：2024/05/17 22:52

http://hihocoder.com/problemset/problem/1401

题意理解：每个人进队离队100次，10000个人，所以用优先队列1e6的nlogn应该可以，而将班级混起来，按照时间t排序就是对的。

急转弯：使用优先队列，班级混起来，用时间t和sno排序来做就是对的：同一个课堂的顺序一定对，而不同课堂，若排队的再其他课堂没排队的前面，那么当前排队的优先级一定也在之后的前面，所以是对的。

算法：无

数据结构：优先队列||堆

#include <iostream>#include <cstdio>#include <queue>#include <cstring>using namespace std;const int maxN = 10000 + 100;const int maxC = 100 + 10;struct node{    int t, s_no, now_id, id;    node(int _t, int _s_no, int _now_id, int _id) {        t = _t;        s_no = _s_no;        now_id = _now_id;        id = _id;    }    bool operator < (const node &o) const {        if(t == o.t) {            return s_no > o.s_no;        }        return t > o.t;    }};priority_queue<node> q;int N, M, K;int c_tim[maxC], p[maxN], ans[maxN];int s[maxN][maxC][2];int main() {    scanf("%d%d%d", &N, &M, &K);    while(!q.empty()) q.pop();    memset(c_tim, 0, sizeof(c_tim));    for(int i = 1; i <= N; i++) {        int si, ti;        scanf("%d%d%d", &si, &ti, &p[i]);        for(int j = 1; j <= p[i]; j++) {            scanf("%d%d", &s[i][j][0], &s[i][j][1]);        }        q.push(node(ti, si, 1, i));    }    while(!q.empty()) {        node u = q.top();        q.pop();        int start_t = max(c_tim[s[u.id][u.now_id][0]], u.t);        int end_t = start_t + s[u.id][u.now_id][1];        c_tim[s[u.id][u.now_id][0]] = end_t;        if(u.now_id == p[u.id]) {            ans[u.id] = end_t;        } else{            q.push(node(end_t + K, u.s_no, u.now_id + 1, u.id));        }    }    for(int i = 1; i <= N; i++) printf("%d\n", ans[i] + K);    return 0;}

python学习使用的heapq，但是python T了。。

from __future__ import print_function##'test for heapq'__author__ = 'hjkruclion'import heapqimport sysdef read_int():     return list(map(int, sys.stdin.readline().split()))maxC = 100 + 10maxN = 10000 + 100N, M, K = read_int()c_tim = [0 for i in range(maxC)]s = [[(0, 0) for j in range(maxC)] for i in range(maxN)]p = [0 for i in range(maxN)]ans = [0 for i in range(maxN)]h = []for i in range(1, N + 1):    t = read_int()    si, ti, p[i] = t[0 : 3]    for j in range(1, p[i] + 1):        s[i][j] = (t[j * 2 - 2 + 3], t[j * 2 + 1 - 2 + 3])    heapq.heappush(h, (ti, si, 1, i))while len(h) > 0:    (t_u, s_no_u, now_id_u, id_u) = heapq.heappop(h)    now_c_id , now_c_tim = s[id_u][now_id_u]    start_t = max(c_tim[now_c_id], t_u)    end_t = start_t + now_c_tim    c_tim[now_c_id] = end_t    if now_id_u == p[id_u]:        ans[id_u] = end_t    else:        get_t = end_t + K        heapq.heappush(h, (get_t, s_no_u, now_id_u + 1, id_u))for i in range(1, N + 1):    print(ans[i] + K)

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