The following code defines a multimap associative container and inserts some ini
ID: 3695784 • Letter: T
Question
The following code defines a multimap associative container and inserts some initial values. Add code for a second multimap associative container and algorithm statements that search for duplicates in the original container and move those duplicate keys and values to the new container (deleting the entries from the original container). See Fig. 15.17 on p. 670.
#include <map>
using namespace std;
int main()
{
multimap< int, int, less< int > > container8;
container8.insert( make_pair ( 7, 24 ) );
container8.insert( make_pair ( 7, 16 ) );
container8.insert( make_pair ( 4, 56 ) );
container8.insert( make_pair ( 4, 72 ) );
container8.insert( make_pair ( 3, 17 ) );
container8.insert( make_pair ( 7, 41 ) );
container8.insert( make_pair ( 2, 12 ) );
container8.insert( make_pair ( 5, 67 ) );
}
Explanation / Answer
#include <iostream>
#include <float.h>
#include <stdlib.h>
#include <math.h>
using namespace std;
// A structure to represent a Point in 2D plane
struct Point
{
int x, y;
};
/* Following two functions are needed for library function qsort().
Refer: http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */
// Needed to sort array of points according to X coordinate
int compareX(const void* a, const void* b)
{
Point *p1 = (Point *)a, *p2 = (Point *)b;
return (p1->x - p2->x);
}
// Needed to sort array of points according to Y coordinate
int compareY(const void* a, const void* b)
{
Point *p1 = (Point *)a, *p2 = (Point *)b;
return (p1->y - p2->y);
}
// A utility function to find the distance between two points
float dist(Point p1, Point p2)
{
return sqrt( (p1.x - p2.x)*(p1.x - p2.x) +
(p1.y - p2.y)*(p1.y - p2.y)
);
}
// A Brute Force method to return the smallest distance between two points
// in P[] of size n
float bruteForce(Point P[], int n)
{
float min = FLT_MAX;
for (int i = 0; i < n; ++i)
for (int j = i+1; j < n; ++j)
if (dist(P[i], P[j]) < min)
min = dist(P[i], P[j]);
return min;
}
// A utility function to find minimum of two float values
float min(float x, float y)
{
return (x < y)? x : y;
}
// A utility function to find the distance beween the closest points of
// strip of given size. All points in strip[] are sorted accordint to
// y coordinate. They all have an upper bound on minimum distance as d.
// Note that this method seems to be a O(n^2) method, but it's a O(n)
// method as the inner loop runs at most 6 times
float stripClosest(Point strip[], int size, float d)
{
float min = d; // Initialize the minimum distance as d
// Pick all points one by one and try the next points till the difference
// between y coordinates is smaller than d.
// This is a proven fact that this loop runs at most 6 times
for (int i = 0; i < size; ++i)
for (int j = i+1; j < size && (strip[j].y - strip[i].y) < min; ++j)
if (dist(strip[i],strip[j]) < min)
min = dist(strip[i], strip[j]);
return min;
}
// A recursive function to find the smallest distance. The array Px contains
// all points sorted according to x coordinates and Py contains all points
// sorted according to y coordinates
float closestUtil(Point Px[], Point Py[], int n)
{
// If there are 2 or 3 points, then use brute force
if (n <= 3)
return bruteForce(Px, n);
// Find the middle point
int mid = n/2;
Point midPoint = Px[mid];
// Divide points in y sorted array around the vertical line.
// Assumption: All x coordinates are distinct.
Point Pyl[mid+1]; // y sorted points on left of vertical line
Point Pyr[n-mid-1]; // y sorted points on right of vertical line
int li = 0, ri = 0; // indexes of left and right subarrays
for (int i = 0; i < n; i++)
{
if (Py[i].x <= midPoint.x)
Pyl[li++] = Py[i];
else
Pyr[ri++] = Py[i];
}
// Consider the vertical line passing through the middle point
// calculate the smallest distance dl on left of middle point and
// dr on right side
float dl = closestUtil(Px, Pyl, mid);
float dr = closestUtil(Px + mid, Pyr, n-mid);
// Find the smaller of two distances
float d = min(dl, dr);
// Build an array strip[] that contains points close (closer than d)
// to the line passing through the middle point
Point strip[n];
int j = 0;
for (int i = 0; i < n; i++)
if (abs(Py[i].x - midPoint.x) < d)
strip[j] = Py[i], j++;
// Find the closest points in strip. Return the minimum of d and closest
// distance is strip[]
return min(d, stripClosest(strip, j, d) );
}
// The main functin that finds the smallest distance
// This method mainly uses closestUtil()
float closest(Point P[], int n)
{
Point Px[n];
Point Py[n];
for (int i = 0; i < n; i++)
{
Px[i] = P[i];
Py[i] = P[i];
}
qsort(Px, n, sizeof(Point), compareX);
qsort(Py, n, sizeof(Point), compareY);
// Use recursive function closestUtil() to find the smallest distance
return closestUtil(Px, Py, n);
}
// Driver program to test above functions
int main()
{
Point P[] = {{2, 3}, {12, 30}, {40, 50}, {5, 1}, {12, 10}, {3, 4}};
int n = sizeof(P) / sizeof(P[0]);
cout << "The smallest distance is " << closest(P, n);
return 0;
}
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