The shortest path of a block

time limit:ms | Memory limit:65535 KB Difficulty:4

Describe

There are many residents in a block, and the streets of the block are only for the east and South directions.

Residents can only walk along the street.

The intervals between the streets are equal.

Use (x, y) to indicate the neighborhood in which the occupants sit.

For example (4,20), represents a user in east-west to 4th Street, North-south direction 20th Street.

Now to build a post office, so that the number of residents to the post office and the minimum distance.

Now this post office should be built in that place to minimize the distance of all households;

Input

The

first line is an integer n<20, which indicates that there are N groups of test data, and the following are N groups of data;
The first line of each group is an integer m<20, indicating that the group has m households, and the following m rows have two integer 0<x,y<100 per row, indicating the coordinates of a user's block.
After M line is a new set of data;

Output

each set of data output to the Post Office minimum distance and, enter the end;

Sample input

231 12 11 252 9 5 2011 91 11 20

Sample output

244

Source

Classic Topics

Uploaded by

Iphxer

//c++

#include <iostream> #include <algorithm>using namespace Std;int x[30],y[30],n,m,i; int main () {cin>>n;while (n--) {cin>>m;for (i=0;i<m;i++) cin>>x[i]>>y[i];sort (x,x+m); sort (y,y+m); int sum=0;for (i=0;i<m/2;i++) Sum+=x[m-1-i]-x[i]+y[m-1-i]-y[i];cout<<sum<<endl;} return 0;}        

//v1.0 C

#include <stdio.h>
#include <stdlib.h>

int main ()
{
int n,m;
scanf ("%d", &n);
GetChar ();
While (n--) {
scanf ("%d", &m);
GetChar ();
int arr[m][2];
int sum = 0;

//input M
for (int i = 0; i < m; ++i) {
For (int j = 0; j < 2; ++j) {
scanf ("%d", &arr[i][j]);
            }
        }


int x=0;
//the bound of X
for (int i = 0; i < m; ++i) {
if (arr[i][0] > x) {
x = arr[i][0];
            }
        }


//the bound of Y
int y=0;
for (int i = 0; i < m; ++i) {
if (arr[i][1] > y) {
y = arr[i][1];
            }
        }



//min Xsum & ysum
int tmp_xmin=2000;
int tmp_ymin=2000;
int p = 0;
int q=0;
While (p<=x) {
int sumx = 0;
for (int i = 0; i < m; ++i) {
sumx + = ABS (arr[i][0]-p);
            }

if (Tmp_xmin > Sumx) {
tmp_xmin = sumx;
            }
p++;
        }

While (q<=y) {
int sumy = 0;
//y
for (int i = 0; i < m; ++i) {
Sumy + = ABS (ARR[I][1]-Q);
            }

if (Tmp_ymin > Sumy) {
tmp_ymin = Sumy;
            }
q++;
        }

sum = tmp_xmin + tmp_ymin;
printf ("%d\n", sum);
    }
return 0;
}

v0.01 C

#include <stdio.h>
#include <stdlib.h>

int main ()
{
int n,m;
scanf ("%d", &n);
GetChar ();
While (n--) {
scanf ("%d", &m);
GetChar ();
int arr[m][2];
int sum = 0;

//input M
for (int i = 0; i < m; ++i) {
For (int j = 0; j < 2; ++j) {
scanf ("%d", &arr[i][j]);
//printf ("\ n");
            }
        }

//output tests
      /*
      *
* for (int i = 0; i < m; ++i) {
For (int j = 0; j < 2; ++j) {
printf ("%d", arr[i][j]);

            }
printf ("\ n");
        }*/


int x=0;
//the bound of X
for (int i = 0; i < m; ++i) {
if (arr[i][0] > x) {
x = arr[i][0];
            }
        }
printf ("Bound x:%d\n", x);

//the bound of Y
int y=0;
for (int i = 0; i < m; ++i) {
if (arr[i][1] > y) {
y = arr[i][1];
            }
        }
printf ("Bound y:%d\n", y);


//min Xsum & ysum
int tmp_xmin=2000;
int tmp_ymin=2000;
int p = 0;

int q =0;
While (p<=x) {
int sumx = 0;
for (int i = 0; i < m; ++i) {
sumx + = ABS (arr[i][0]-p);
            }
if (Tmp_xmin > Sumx) {
tmp_xmin = sumx;
            }
p++;
        }

while (q<=y) {
int sumy = 0;
for (int i = 0; i < m; ++i) {
Sumy + = ABS (ARR[I][0]-Q);
}

Y
for (int i = 0; i < m; ++i) {
Sumy + = ABS (ARR[I][1]-P);
}
if (Tmp_ymin > Sumy) {
Tmp_ymin = Sumy;
}
q++;
}

printf ("tmp_xmin:%d\n", tmp_xmin);
printf ("tmp_ymin:%d\n", tmp_ymin);

sum = tmp_xmin + tmp_ymin;
printf ("summin:%d\n", sum);
}
return 0;
}

The shortest path of a block