Hdu 4311-Meeting point-1 曼哈顿距离,前缀和

题目：http://acm.hdu.edu.cn/showproblem.php?pid=4311

Meeting point-1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3426    Accepted Submission(s): 1131

Problem Description

It has been ten years since TJU-ACM established. And in this year all the retired TJU-ACMers want to get together to celebrate the tenth anniversary. Because the retired TJU-ACMers may live in different places around the world, it may be hard to find out where to celebrate this meeting in order to minimize the sum travel time of all the retired TJU-ACMers. 
There is an infinite integer grid at which N retired TJU-ACMers have their houses on. They decide to unite at a common meeting place, which is someone's house. From any given cell, only 4 adjacent cells are reachable in 1 unit of time.
Eg: (x,y) can be reached from (x-1,y), (x+1,y), (x, y-1), (x, y+1).
Finding a common meeting place which minimizes the sum of the travel time of all the retired TJU-ACMers.

 

Input

The first line is an integer T represents there are T test cases. (0<T <=10)
For each test case, the first line is an integer n represents there are n retired TJU-ACMers. (0<n<=100000), the following n lines each contains two integers x, y coordinate of the i-th TJU-ACMer. (-10^9 <= x,y <= 10^9)

 

Output

For each test case, output the minimal sum of travel times.

 

Sample Input

4 6 -4 -1 -1 -2 2 -4 0 2 0 3 5 -2 6 0 0 2 0 -5 -2 2 -2 -1 2 4 0 5 -5 1 -1 3 3 1 3 -1 1 -1 10 -1 -1 -3 2 -4 4 5 2 5 -4 3 -1 4 3 -1 -2 3 4 -2 2

 

Sample Output

26 20 20 56

Hint

In the first case, the meeting point is (-1,-2); the second is (0,0), the third is (3,1) and the last is (-2,2)

 

Author

TJU

 

Source

2012 Multi-University Training Contest 2 

 

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题意：给你n个点的坐标，让你找其中一个坐标，使得所有点到那个点的曼哈顿距离和最小。

题解：

曼哈顿距离＋前缀和

把x和y分别排序，然后去维护分别排序后的前缀和，还有前缀和的前缀和。

注意开long long

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100010
#define INF 10000000000000000LL
#define LL long long
LL qx[MAXN],qy[MAXN],qx1[MAXN],qy1[MAXN];
LL ans[MAXN];
struct node
{
    int a,id;
}x[MAXN],y[MAXN];
int read()
{
    int s=0,fh=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();}
    return s*fh;
}
bool cmp(node aa,node bb)
{
    return aa.a<bb.a;
}
int main()
{
    int T,n,i;
    LL MN,sum,sum1;
    T=read();
    while(T--)
    {
        n=read();
        for(i=1;i<=n;i++){x[i].a=read(),y[i].a=read();x[i].id=i;y[i].id=i;}
        sort(x+1,x+n+1,cmp);
        sort(y+1,y+n+1,cmp);
        qx[1]=0;qy[1]=0;qx1[1]=0;qy1[1]=0;
        for(i=2;i<=n;i++)
        {
            qx[i]=qx[i-1]+(LL)(x[i].a-x[i-1].a);
            qx1[i]=qx1[i-1]+qx[i];
            qy[i]=qy[i-1]+(LL)(y[i].a-y[i-1].a);
            qy1[i]=qy1[i-1]+qy[i];
        }
        memset(ans,0,sizeof(ans));
        for(i=1;i<=n;i++)
        {
            sum=qx[i]*(i-1)-qx1[i-1]+qx1[n]-qx1[i]-qx[i]*(n-i);
            sum1=qy[i]*(i-1)-qy1[i-1]+qy1[n]-qy1[i]-qy[i]*(n-i);
            ans[x[i].id]+=sum;
            ans[y[i].id]+=sum1;
        }
        MN=INF;
        for(i=1;i<=n;i++)MN=min(MN,ans[i]);
        printf("%lld
",MN);
    }
    return 0;
}