codeforces C. Queue (思维)

本文介绍了一种解决特定队列问题的方法,该问题要求根据每个人前面比自己高的人数来还原队列顺序,并给出每个人的可能身高。通过排序和逐个调整高度的方式解决了这一挑战。

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C. Queue
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

In the Main Berland Bank n people stand in a queue at the cashier, everyone knows his/her height hi, and the heights of the other people in the queue. Each of them keeps in mind number ai — how many people who are taller than him/her and stand in queue in front of him.

After a while the cashier has a lunch break and the people in the queue seat on the chairs in the waiting room in a random order.

When the lunch break was over, it turned out that nobody can remember the exact order of the people in the queue, but everyone remembers his number ai.

Your task is to restore the order in which the people stood in the queue if it is possible. There may be several acceptable orders, but you need to find any of them. Also, you need to print a possible set of numbers hi — the heights of people in the queue, so that the numbers ai are correct.

Input

The first input line contains integer n — the number of people in the queue (1 ≤ n ≤ 3000). Then n lines contain descriptions of the people as "namei ai" (one description on one line), where namei is a non-empty string consisting of lowercase Latin letters whose length does not exceed 10 characters (the i-th person's name), ai is an integer (0 ≤ ai ≤ n - 1), that represents the number of people who are higher and stand in the queue in front of person i. It is guaranteed that all names are different.

Output

If there's no acceptable order of the people in the queue, print the single line containing "-1" without the quotes. Otherwise, print in n lines the people as "namei hi", where hi is the integer from 1 to 109(inclusive), the possible height of a man whose name is namei. Print the people in the order in which they stand in the queue, starting from the head of the queue and moving to its tail. Numbers hi are not necessarily unique.

Examples
input
4
a 0
b 2
c 0
d 0
output
a 150
c 170
d 180
b 160
input
4
vasya 0
petya 1
manya 3
dunay 3
output
-1

知道每个人前 面比自己高的人的个数,先从小到大排序,先确定i自己的身高的大小(就是自己再人群里身高排第几),再把序列中1~i-1>=i.h的高度都加1

code:

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct Node
{
    int num;
    char str[20];
    int a;
    int h;
}p[3010];
bool cmp(Node x,Node y)
{
    return x.a<y.a;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        cin>>p[i].str>>p[i].a;
        p[i].num=i;
    }
    sort(p+1,p+1+n,cmp);
    int flag=0;
    for(int i=1;i<=n;i++)
    {
        if(p[i].a>=i)
        {
            flag=1;
            break;
        }
        p[i].h=i-p[i].a;
        for(int j=1;j<i;j++)
        {
            if(p[j].h>=p[i].h)
                p[j].h++;
        }
    }
    if(flag)
        printf("-1\n");
    else
    {
        for(int i=1;i<=n;i++)
        {
            printf("%s %d\n",p[i].str,p[i].h);
        }
    }
    return 0;
}


### Codeforces Div.2 比赛难度介绍 Codeforces Div.2 比赛主要面向的是具有基础编程技能到中级水平的选手。这类比赛通常吸引了大量来自全球不同背景的参赛者,包括大学生、高中生以及一些专业人士。 #### 参加资格 为了参加 Div.2 比赛,选手的评级应不超过 2099 分[^1]。这意味着该级别的竞赛适合那些已经掌握了一定算法知识并能熟练运用至少一种编程语言的人群参与挑战。 #### 题目设置 每场 Div.2 比赛一般会提供五至七道题目,在某些特殊情况下可能会更多或更少。这些题目按照预计解决难度递增排列: - **简单题(A, B 类型)**: 主要测试基本的数据结构操作和常见算法的应用能力;例如数组处理、字符串匹配等。 - **中等偏难题(C, D 类型)**: 开始涉及较为复杂的逻辑推理能力和特定领域内的高级技巧;比如图论中的最短路径计算或是动态规划入门应用实例。 - **高难度题(E及以上类型)**: 对于这些问题,则更加侧重考察深入理解复杂概念的能力,并能够灵活组合多种方法来解决问题;这往往需要较强的创造力与丰富的实践经验支持。 对于新手来说,建议先专注于理解和练习前几类较容易的问题,随着经验积累和技术提升再逐步尝试更高层次的任务。 ```cpp // 示例代码展示如何判断一个数是否为偶数 #include <iostream> using namespace std; bool is_even(int num){ return num % 2 == 0; } int main(){ int number = 4; // 测试数据 if(is_even(number)){ cout << "The given number is even."; }else{ cout << "The given number is odd."; } } ```
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