codeforces Inna and Alarm Clock

本博客探讨了一个问题,即如何通过最少的垂直或水平线段移动,来关闭房间内所有放置闹钟的位置。具体步骤包括选择段类型、进行多次移动,最终目标是高效地关闭所有闹钟。

Description

Inna loves sleeping very much, so she needs n alarm clocks in total to wake up. Let’s suppose that Inna’s room is a 100 × 100 square with the lower left corner at point (0, 0) and with the upper right corner at point (100, 100). Then the alarm clocks are points with integer coordinates in this square.

The morning has come. All n alarm clocks in Inna’s room are ringing, so Inna wants to turn them off. For that Inna has come up with an amusing game:
• First Inna chooses a type of segments that she will use throughout the game. The segments can be either vertical or horizontal.
• Then Inna makes multiple moves. In a single move, Inna can paint a segment of any length on the plane, she chooses its type at the beginning of the game (either vertical or horizontal), then all alarm clocks that are on this segment switch off. The game ends when all the alarm clocks are switched off.

Inna is very sleepy, so she wants to get through the alarm clocks as soon as possible. Help her, find the minimum number of moves in the game that she needs to turn off all the alarm clocks!

Input

The first line of the input contains integer n(1 ≤ n ≤ 105) — the number of the alarm clocks. The next n lines describe the clocks: the i-th line contains two integers xi, yi — the coordinates of the i-th alarm clock (0 ≤ xi, yi ≤ 100).

Note that a single point in the room can contain any number of alarm clocks and the alarm clocks can lie on the sides of the square that represents the room.

Output

In a single line print a single integer — the minimum number of segments Inna will have to draw if she acts optimally.

Sample Input

Input
4
0 0
0 1
0 2
1 0

Output
2

Input
4
0 0
0 1
1 0
1 1

Output
2

Input
4
1 1
1 2
2 3
3 3

Output
3

Hint

In the first sample, Inna first chooses type “vertical segments”, and then she makes segments with ends at : (0, 0), (0, 2); and, for example, (1, 0), (1, 1). If she paints horizontal segments, she will need at least 3 segments.

In the third sample it is important to note that Inna doesn’t have the right to change the type of the segments during the game. That’s why she will need 3 horizontal or 3 vertical segments to end the game.

题意:输入n,接着是n个坐标;判断给定数据有几个不同的行,有几个不同的列,输出较小值;

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int s[105][105];
int max(int x,int y)
{
    return x>y?x:y;
}
int min(int x,int y)
{
    return x<y?x:y;
}
int main()
{
    int n,f=0;
    cin>>n;
    memset(s,0,sizeof(s));
    for(int i=0; i<n; i++)
    {
        int a,b;
        cin>>a>>b;
        f=max(a,f);
        f=max(f,b);
        s[a][b]=1;
    }
    int s1=0,s2=0;
    for(int i=0; i<=f; i++)
    {
        for(int j=0; j<=f; j++)
        {
            if(s[i][j]==1)
            {
                s1++;
                break;
            }
        }
    }
    for(int i=0; i<=f; i++)
    {
        for(int j=0; j<=f; j++)
        {
            if(s[j][i]==1)
            {
                s2++;
                break;
            }
        }
    }
    s1=min(s1,s2);
    cout<<s1<<endl;
    return 0;
}
虽然给定引用中未直接提及“Kuroni and Simple Strings”题目的详细信息,但通常这类题目可能与字符串处理、括号匹配等相关。一般而言,题目可能会给出一个由括号组成的字符串,要求找出能移除的最大数量的不相交的合法括号对,并输出移除这些括号对后的相关信息。 ### 解法分析 #### 栈解法 栈解法是处理括号匹配问题的经典方法。通过遍历字符串,将左括号压入栈中,遇到右括号时,若栈顶为左括号,则将栈顶元素弹出,表示这是一对匹配的括号。 ```python s = input() stack = [] pairs = [] for i, char in enumerate(s): if char == '(': stack.append(i) else: if stack: left_index = stack.pop() pairs.append((left_index + 1, i + 1)) if not pairs: print(0) else: print(1) print(len(pairs) * 2) result = [] for l, r in pairs: result.extend([l, r]) result.sort() print(" ".join(map(str, result))) ``` #### 双指针解法 双指针解法从字符串的两端向中间遍历,分别使用两个指针 `left` 和 `right`。`left` 指针从左向右寻找 `(`,`right` 指针从右向左寻找 `)`,当找到一对匹配的括号时,将它们标记为已移除,继续寻找下一对匹配的括号,直到无法再找到匹配的括号为止。 ```python s = input() n = len(s) left = 0 right = n - 1 pairs = [] while left < right: while left < right and s[left] != '(': left += 1 while left < right and s[right] != ')': right -= 1 if left < right: pairs.append((left + 1, right + 1)) left += 1 right -= 1 if not pairs: print(0) else: print(1) print(len(pairs) * 2) result = [] for l, r in pairs: result.extend([l, r]) result.sort() print(" ".join(map(str, result))) ``` ### 复杂度分析 - **栈解法**:时间复杂度为 $O(n)$,其中 $n$ 是字符串的长度。空间复杂度为 $O(n)$,主要用于栈的空间开销。 - **双指针解法**:时间复杂度为 $O(n)$,空间复杂度为 $O(n)$,主要用于存储匹配的括号对。
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