HDU4311 Meeting point-1 (Manhattan distance), hdu4311point-1

HDU4311 Meeting point-1 (Manhattan distance), hdu4311point-1

Question link:

Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 4311

Question:

Given n points, select one of them as the starting point, and then minimize the distance from other points to the Manhattan of this point.

Analysis:

Let's set P as the starting point.

Then ans = sum (abs | pi. x-p.x | + | pi. y-p.y |) (1 <= I <= n)

We can sort them by x and y respectively to remove the absolute value symbol.

Then the simplified formula can be changed

Set this point to I after sorting by x;

If tot [I] is set, it indicates the sum of the abscissa of the point to sequence I.

Ansx = (I-1) * p. x + (tot [n]-tot [I])-(n-I) * p. x

Similarly, we can find

Set this point to I after sorting by t;

Ansy = (I-1) * p. y + (tot [n]-tot [I])-(n-I) * p. y

Ans = ansx + ansy

Take the smallest ans.

The time complexity is O (nlog (n ));

The Code is as follows:

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 1e5+10;struct point{    int x,y;    LL sum;}p[maxn];bool cmp1(point A,point B){    if(A.x<B.x) return true;    if(A.x==B.x&&A.y<B.y) return true;    return false;}bool cmp2(point A,point B){    if(A.y<B.y) return true;    if(A.y==B.y&&A.y<B.y) return true;    return false;}int main(){    int t,n;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        LL sum=0;        for(int i=1;i<=n;i++){            scanf("%d%d",&p[i].x,&p[i].y);            p[i].sum=0;        }        sort(p+1, p+1+n, cmp1);        for (LL i = 1; i <= n; ++i) {            p[i].sum += (i-1) * p[i].x - sum;            sum += p[i].x;        }        sum = 0;        for (LL i = n; i >= 1; --i) {            p[i].sum += sum - (n-i) * p[i].x;            sum += p[i].x;        }        sum = 0;        sort(p+1, p+1+n, cmp2);        for (LL i = 1; i <= n; ++i) {            p[i].sum += (i-1) * p[i].y -sum;            sum += p[i].y;        }        sum = 0;        LL ans = 1LL<<62;        for (LL i = n; i >= 1; --i) {            p[i].sum += sum - (n-i) * p[i].y;            ans = min(ans, p[i].sum);            sum += p[i].y;        }        printf("%I64d\n",ans);    }    return 0;}

Zookeeper