2018今日頭條大數據方向筆試題

題目鏈接

一、一道繁瑣的大模擬

N個產品經理，M個程序員，P個需求。N個產品經理向M個程序員提出P個需求。
每個需求用四元組表示：產品經理ID（誰提出的） 提出時間 優先級（數字越大優先級越高） 需求開發時長
每個產品經理對於自己的需求按照：優先級從高到低、開發時長從短到廠、提出時間 從早到晚排序
每個程序員每次從產品經理最想做的需求里面挑選用時最短（如果時間相同優先做PM編號較小的需求）

求每個需求的完成時間。

import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
class Idea {
    int pm, think, use, pri, index;
    int over;

    Idea(int pm, int think, int pri, int use, int index) {
        this.pm = pm;
        this.think = think;
        this.use = use;
        this.pri = pri;
        this.index = index;
    }

    @Override
    public String toString() {
        return "pm:" + pm + " think:" + think + " use" + use + " pri:" + pri;
    }
}

class Pm {
    PriorityQueue<Idea> ideas = new PriorityQueue<>((x, y) -> {
        if (x.pri != y.pri) return -x.pri + y.pri;
        if (x.use != y.use) return x.use - y.use;
        return x.think - y.think;
    });

    boolean want(Idea i) {
        return ideas.peek() == i;
    }
}

class Programmer {
    int time = 0;
}

Main() {
    Scanner cin = new Scanner(System.in);
    int PMCount = cin.nextInt(), ProgrammerCouunt = cin.nextInt(), IdeaCount = cin.nextInt();
    Idea[] a = new Idea[IdeaCount];
    Pm pms[] = new Pm[PMCount + 1];
    for (int i = 0; i <= PMCount; i++) {
        pms[i] = new Pm();
    }
    for (int i = 0; i < IdeaCount; i++) {
        a[i] = new Idea(cin.nextInt(), cin.nextInt(), cin.nextInt(), cin.nextInt(), i);
    }
    PriorityQueue<Programmer> consumer = new PriorityQueue<>(Comparator.comparing(x -> x.time));
    Arrays.sort(a, Comparator.comparingInt(x -> x.think));
    int ai = 0;
    PriorityQueue<Idea> producer = new PriorityQueue<>((x, y) -> {
        if (x.use != y.use) return x.use - y.use;
        return x.pm - y.pm;
    });
    for (int i = 0; i < ProgrammerCouunt; i++) consumer.add(new Programmer());
    while (!producer.isEmpty() || ai < a.length) {
        Programmer worker = consumer.poll();
        if (producer.isEmpty()) {
            worker.time = Math.max(a[ai].think, worker.time);
        }
        while (ai < a.length && a[ai].think <= worker.time) {
            Idea task = a[ai++];
            pms[task.pm].ideas.add(task);
            producer.add(task);
        }
        Idea idea = producer.poll();
        Pm pm = pms[idea.pm];
        if (pm.want(idea)) {//如果當前idea是pm最想做的idea
            pm.ideas.poll();//pm完成了這個任務
            if (!pm.ideas.isEmpty()) producer.add(pm.ideas.peek());
            worker.time = idea.over = Math.max(worker.time, idea.think) + idea.use;
//            System.out.println("idea " + idea + " " + worker.time);
        }
        consumer.add(worker);
    }
    Arrays.sort(a, Comparator.comparingInt(x -> x.index));
    for (Idea i : a) {
        System.out.println(i.over);
    }
}

public static void main(String[] args) {
    new Main();
}
}

二、單調棧的變形應用

給定數組a[N]，它有許多區間，求s=min(a[l:r]) \times sum(a[l:r])的最大值。a[N]中全為正整數。

sum(a[l:r])可以快速通過前綴和求出，因為要求s的最大值，所以區間[l:r]應該盡量大。但是又不能太大，因為太大的話，會使得min(a[l:r])很小。
對於a中的每個元素a[i]，在保證a[i]為最小值得情況下，擴張左區間和右區間。問題轉化為快速求解a[i]左右第一個比它小的元素。

#include<iostream>
#include<stdio.h>
#include<map>
#include<algorithm>
#include<stack>
using namespace std;
const int maxn = 5 * 1e5;
int n;
int a[maxn];
int s[maxn];
int lef[maxn], righ[maxn];//每個數字左邊比我小的和右邊比我小的
int getsum(int f, int t) {
    if (t == 0)return s[f];
    return s[t] - s[f - 1];
}
int main() {
    cin >> n;
    for (int i = 0; i < n; i++)scanf("%d", a + i);
    int nows = 0;
    for (int i = 0; i < n; i++) {
        s[i] = nows + a[i];
        nows += a[i];
    }
    stack<int>sta;
    for (int i = 0; i < n; i++) {
        while (!sta.empty()) {
            int topIndex = sta.top();
            int value = a[topIndex];
            if (value > a[i]) {
                sta.pop();
                righ[topIndex] = i - 1;
            }
            else {
                break;
            }
        }
 
        int last = (sta.empty() ? 0 : sta.top() + 1);
        lef[i] = last;
        sta.push(i);
    }
    while (!sta.empty()) {
        int topIndex = sta.top();
        sta.pop();
        righ[topIndex] = n - 1;
    }
    long long  s = 0;
    for (int i = 0; i < n; i++) {
        s = max(s, getsum(lef[i], righ[i])*(long long)a[i]);
    }
    cout << s << endl;
    return 0;
}

三、求邊界點

平面直角坐標系第一象限中包含一堆整數點，請找到全部滿足以下條件的點：(x,y)的左方、上方一個點也沒有。
這道題簡單的排序就能通過，但是因為數據量太大，用Java很容易超時，用C++ cin都會超時，必須用scanf才可以。

#include<iostream>
#include<stdio.h>
#include<map>
#include<algorithm>
using namespace std;
const int maxn = 5 * 1e5;
int n;
struct Point {
    int x, y;
}a[maxn];
int b[maxn];
int bi;
bool com(const Point&one, const Point&two) {
    if (one.x != two.x)return one.x<two.x;
    else return one.y < two.y;
}
int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        scanf("%d%d", &a[i].x, &a[i].y);
    }
    sort(a, a + n, com);
    bi = 0;
    for (int i = 0; i < n; i++) {
        if (i == n || a[i].x != a[i + 1].x) {
            b[bi++] = i;
        }
    }
    int ma = 0;
    for (int i = bi - 1; i >= 0; i--) {
        int ind = b[i];
        if (ma > a[ind].y) {
            b[i] = -1;
        }
        ma = max(ma, a[ind].y);
    }
    for (int i = 0; i < bi; i++) {
        if (b[i] != -1) {
            int ind = b[i];
            cout << a[ind].x << " " << a[ind].y << endl;
        }
    }
    return 0;
}