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📜  检查是否可以将框连接成一个圆圈的查询

📅  最后修改于: 2021-10-27 08:53:11             🧑  作者: Mango

考虑按顺时针方向(从 1 到 n 编号)按递增顺序排列在圆圈中的n 个盒子。给你q 个查询,每个查询包含两个整数 i 和 j。任务是检查是否有可能通过一根杆将框 i连接到框 j而不与之前查询中用于连接其他框的杆相交。此外,每个盒子最多可以连接到另一个盒子,而没有盒子可以连接到它自己。
例子:

方法:
假设盒子 x 已经连接到盒子 y。我们需要将盒子 i 连接到盒子 j。
现在,观察可以有两种情况,连接盒 i 和盒 j 的两个杆会与连接杆连接盒 x 和盒 y 相交。
情况 1: x < i 并且 y 位于 i 和 j 之间:

情况 2: x 位于 i 和 j 之间且 y >j:

我们将明确检查杆是否打算连接到自身,或者杆是否打算连接两个盒子,使得其中至少一个已经连接的情况。
因此,我们将检查上述两个条件。如果两个中的任何一个相遇,则无法连接,否则我们可以连接盒子。

下面是上述方法的实现:

C++
// C++ implementation of above approach
#include 
using namespace std;
#define MAX 50
 
// Print the answer to each query
void solveQuery(int n, int q, int qi[], int qj[])
{
    int arr[MAX];
    for (int i = 0; i <= n; i++)
        arr[i] = 0;
 
    for (int k = 0; k < q; k++) {
 
        // setting the flag for exception
        int flag = 0;
 
        // replacing the greater element in i and j
        if (qj[k] < qi[k]) {
            int temp = qi[k];
            qi[k] = qj[k];
            qj[k] = temp;
        }
 
        // checking if that box is not
        // used in previous query.
        if (arr[qi[k]] != 0 || arr[qj[k]] != 0)
            flag = 1;
 
        // checking if connecting to the same box
        else if (qi[k] == qj[k])
            flag = 1;
 
        else {
 
            // case 1: x < i and y lies between i and j
            for (int i = 1; i < qi[k]; i++) {
                if (arr[i] != 0 && arr[i] < qj[k] && qi[k] < arr[i]) {
                    flag = 1;
                    break;
                }
            }
 
            // case 2: x lies between i and j and y >j
            if (flag == 0) {
                for (int i = qi[k] + 1; i < qj[k]; i++) {
                    if (arr[i] != 0 && arr[i] > qj[k]) {
                        flag = 1;
                        break;
                    }
                }
            }
        }
 
        // if flag is not reset inbetween.
        if (flag == 0) {
            cout << "YES\n";
            arr[qi[k]] = qj[k];
            arr[qj[k]] = qi[k];
        }
        else
            cout << "NO\n";
    }
}
 
// Driver code
int main()
{
    int n = 10;
    int q = 7;
    int qi[] = { 1, 2, 2, 2, 9, 10, 8 };
    int qj[] = { 5, 7, 3, 4, 9, 9, 6 };
 
    solveQuery(n, q, qi, qj);
 
    return 0;
}


Java
// Java implementation of
// above approach
class GFG
{
static int MAX = 50;
 
// Print the answer to each query
static void solveQuery(int n, int q,
                       int qi[], int qj[])
{
int[] arr = new int[MAX];
for (int i = 0; i <= n; i++)
    arr[i] = 0;
 
for (int k = 0; k < q; k++)
{
 
    // setting the flag for exception
    int flag = 0;
 
    // replacing the greater
    // element in i and j
    if (qj[k] < qi[k])
    {
        int temp = qi[k];
        qi[k] = qj[k];
        qj[k] = temp;
    }
 
    // checking if that box is not
    // used in previous query.
    if (arr[qi[k]] != 0 ||
        arr[qj[k]] != 0)
        flag = 1;
 
    // checking if connecting
    // to the same box
    else if (qi[k] == qj[k])
        flag = 1;
 
    else
    {
 
        // case 1: x < i and y lies
        // between i and j
        for (int i = 1; i < qi[k]; i++)
        {
            if (arr[i] != 0 && arr[i] < qj[k] &&
                                qi[k] < arr[i])
            {
                flag = 1;
                break;
            }
        }
 
        // case 2: x lies between
        // i and j and y >j
        if (flag == 0)
        {
            for (int i = qi[k] + 1;
                     i < qj[k]; i++)
            {
                if (arr[i] != 0 && arr[i] > qj[k])
                {
                    flag = 1;
                    break;
                }
            }
        }
    }
 
    // if flag is not reset inbetween.
    if (flag == 0)
    {
        System.out.println("YES");
        arr[qi[k]] = qj[k];
        arr[qj[k]] = qi[k];
    }
    else
        System.out.println("NO");
}
}
 
// Driver code
public static void main(String[] args)
{
    int n = 10;
    int q = 7;
    int qi[] = { 1, 2, 2, 2, 9, 10, 8 };
    int qj[] = { 5, 7, 3, 4, 9, 9, 6 };
 
    solveQuery(n, q, qi, qj);
}
}
 
// This code is contributed
// by ChitraNayal


Python 3
# Python 3 implementation of
# above approach
 
MAX = 50
 
# Print the answer to each query
def solveQuery(n, q, qi, qj):
 
    arr = [None] * MAX
    for i in range(n + 1):
        arr[i] = 0
 
    for k in range(q):
 
        # setting the flag
        # for exception
        flag = 0
 
        # replacing the greater
        # element in i and j
        if (qj[k] < qi[k]):
            qj[k], qi[k] = qi[k], qj[k]
 
        # checking if that box is not
        # used in previous query.
        if (arr[qi[k]] != 0 or
            arr[qj[k]] != 0):
            flag = 1
 
        # checking if connecting
        # to the same box
        elif (qi[k] == qj[k]):
            flag = 1
 
        else :
 
            # case 1: x < i and y
            # lies between i and j
            for i in range(1, qi[k]) :
                if (arr[i] != 0 and
                    arr[i] < qj[k] and
                    qi[k] < arr[i]):
                    flag = 1
                    break
 
            # case 2: x lies between
            # i and j and y >j
            if (flag == 0):
                for i in range(qi[k] + 1, qj[k]) :
                    if (arr[i] != 0 and
                        arr[i] > qj[k]):
                        flag = 1
                        break
                     
        # if flag is not reset inbetween.
        if (flag == 0):
            print("YES")
            arr[qi[k]] = qj[k]
            arr[qj[k]] = qi[k]
         
        else:
            print("NO")
 
# Driver code
if __name__ == "__main__":
    n = 10
    q = 7
    qi = [ 1, 2, 2, 2, 9, 10, 8 ]
    qj = [ 5, 7, 3, 4, 9, 9, 6 ]
 
    solveQuery(n, q, qi, qj)
 
# This code is contributed
# by ChitraNayal


C#
// C# implementation of
// above approach
using System;
 
class GFG
{
 
static int MAX = 50;
 
// Print the answer to each query
static void solveQuery(int n, int q,
                       int[] qi, int[] qj)
{
int[] arr = new int[MAX];
for (int i = 0; i <= n; i++)
    arr[i] = 0;
 
for (int k = 0; k < q; k++)
{
 
    // setting the flag for exception
    int flag = 0;
 
    // replacing the greater
    // element in i and j
    if (qj[k] < qi[k])
    {
        int temp = qi[k];
        qi[k] = qj[k];
        qj[k] = temp;
    }
 
    // checking if that box is not
    // used in previous query.
    if (arr[qi[k]] != 0 || arr[qj[k]] != 0)
        flag = 1;
 
    // checking if connecting
    // to the same box
    else if (qi[k] == qj[k])
        flag = 1;
 
    else
    {
 
        // case 1: x < i and y lies
        // between i and j
        for (int i = 1; i < qi[k]; i++)
        {
            if (arr[i] != 0 && arr[i] < qj[k] &&
                                qi[k] < arr[i])
            {
                flag = 1;
                break;
            }
        }
 
        // case 2: x lies between
        // i and j and y >j
        if (flag == 0)
        {
            for (int i = qi[k] + 1;
                     i < qj[k]; i++)
            {
                if (arr[i] != 0 &&
                    arr[i] > qj[k])
                {
                    flag = 1;
                    break;
                }
            }
        }
    }
 
    // if flag is not reset inbetween.
    if (flag == 0)
    {
        Console.Write("YES\n");
        arr[qi[k]] = qj[k];
        arr[qj[k]] = qi[k];
    }
    else
        Console.Write("NO\n");
}
}
 
// Driver code
public static void Main()
{
    int n = 10;
    int q = 7;
    int[] qi = { 1, 2, 2, 2, 9, 10, 8 };
    int[] qj = { 5, 7, 3, 4, 9, 9, 6 };
 
    solveQuery(n, q, qi, qj);
}
}
 
// This code is contributed
// by ChitraNayal


PHP
j
            if ($flag == 0)
            {
                for ($i = $qi[$k] + 1;
                     $i < $qj[$k]; $i++)
                {
                    if ($arr[$i] != 0 &&
                        $arr[$i] > $qj[$k])
                    {
                        $flag = 1;
                        break;
                    }
                }
            }
        }
 
        // if flag is not reset inbetween.
        if ($flag == 0)
        {
            echo "YES\n";
            $arr[$qi[$k]] = $qj[$k];
            $arr[$qj[$k]] = $qi[$k];
        }
        else
            echo "NO\n";
    }
}
 
// Driver code
$n = 10;
$q = 7;
$qi = array( 1, 2, 2, 2, 9, 10, 8 );
$qj = array( 5, 7, 3, 4, 9, 9, 6 );
 
solveQuery($n, $q, $qi, $qj);
 
// This code is contributed
// by ChitraNayal
?>


Javascript


输出:
YES
NO
YES
NO
NO
YES
YES

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