水题：meeting point-1

Meeting point-1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 987    Accepted Submission(s): 306

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

46-4 -1-1 -22 -40 20 35 -260 02 0-5 -22 -2-1 24 05-5 1-1 33 13 -11 -110-1 -1-3 2-4 45 25 -43 -14 3-1 -23 4-2 2

 

Sample Output

26202056
Hint

平面上两点间的 Manhattan 距离为 |x1-x2| + |y1-y2|

X 方向的距离与 Y 方向上的距离可以分开来处理。假设我们以 (xi,yi) 作为开会的地点，那么其余的点到该开会地点所需的时间为 X 方向上到 xi 所需要的时间加上 Y 方向上到 yi 所需要的时间。

对数据预处理后可以快速地求出x坐标小于xi的点的个数rankx, 并且这些 x 坐标值之和 sumx，那么这些点 X 方向上对结果的贡献为 rankx * xi - sumx；同理可以处理出 x 坐标大于 xi 的点在 X 方向上对结果的贡献值。同理可求得其余点在 Y 方向上到达 yi 所需要的总时间。

 

注意INF的值要取足够大，不然不停地wa，最后发现取1LL<<63-1就行了

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>using namespace std;const long long INF =  1LL<<63-1;struct acmers{    long long x;    long long y;    long long sum;}a[100010];bool cmp1(acmers a, acmers b){    return a.x < b.x;}bool cmp2(acmers a, acmers b){    return a.y < b.y;}int main(){    long long t;    long long i;    long long n;    long long rankx, ranky, rankxr, rankyd;    long long sumxl, sumxr;    long long sumyu, sumyd;    scanf("%I64d", &t);    while(t --){        scanf("%I64d", &n);        sumxr = 0;        sumyd = 0;        for(i = 0; i < n; i ++){            scanf("%I64d %I64d", &a[i].x, &a[i].y);            a[i].sum = 0;            sumxr += a[i].x;            sumyd += a[i].y;        }        //cout << "sumxr" << sumxr << endl;        //cout << "sumyd" << sumyd << endl;        sort(a, a+n, cmp1);        sumxl = 0;        rankx = 0;        rankxr = n;        for(rankx = 0; rankx < n; rankx ++){            sumxr -= a[rankx].x;            rankxr--;            a[rankx].sum += rankx * a[rankx].x - sumxl + sumxr - rankxr * a[rankx].x;            sumxl += a[rankx].x;        }        sort(a, a + n, cmp2);        sumyu = 0;        ranky = 0;        rankyd = n;        for(ranky = 0; ranky < n; ranky ++){            sumyd -= a[ranky].y;            rankyd--;            a[ranky].sum += ranky * a[ranky].y - sumyu + sumyd - rankyd * a[ranky].y;            sumyu += a[ranky].y;        }        long long mm = INF;        for(i = 0; i < n; i ++){            if (a[i].sum < mm)                mm = a[i].sum;        }        cout << mm << endl;    }    return 0;}