HDU--4312（坐标变换，切比雪夫距离）

2015-05-16 01:22:40

题目：跟 4311 题意差不多，给出N个点（1e5），找出一个点，使得这个点到其他所有点的切比雪夫距离和最小。（切比雪夫距离即：max（x 坐标差的绝对值，y 坐标差的绝对值 ））

思路：切比雪夫距离和棋盘很有关系，鉴于八皇后中的 x+y 表示法，我们

　　可以想到将每个点的坐标进行变换，(x,y) -> (x+y,x-y)，这样一来，计算两点间的曼哈顿距离

　　就等于两点间切比雪夫距离的两倍。

　　|x1-x2| + |y1-y2|  --->  |(x1+y1)-(x2+y2)| + |(x1-y1)-(x2-y2)| = max(|x1-x2| , |y1-y2|) 

　　注意：可先将每个点的坐标 +2e9 以消除负数。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

#define getmid(l,r) ((l) + ((r) - (l)) / 2)
#define MP(a,b) make_pair(a,b)
#define PB(a) push_back(a)

typedef long long ll;
typedef pair<int,int> pii;
const double eps = 1e-8;
const int INF = (1 << 30) - 1;
const int MAXN = 100010;
const ll RA1 = 2e9;
const ll RA2 = 4e9;

ll sumx[MAXN],sumy[MAXN];
int n;

struct point{
    ll x,y;
    int id;
}p[MAXN];

bool cmpx(point a,point b){ return a.x < b.x; }
bool cmpy(point a,point b){ return a.y < b.y; }
bool cmpid(point a,point b){ return a.id < b.id; }

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i = 1; i <= n; ++i){
            scanf("%lld%lld",&p[i].x,&p[i].y);
            p[i].x = p[i].x + p[i].y;
            p[i].y = p[i].x - p[i].y - p[i].y;
            p[i].x += RA1;
            p[i].y += RA1;
            p[i].id = i;
        }
        sort(p + 1,p + n + 1,cmpx);
        ll sum = 0;
        for(int i = 1; i <= n; ++i){
            int id = p[i].id;
            sum += p[i].x;
            sumx[id] = p[i].x * (ll)i - sum;
        }
        sum = 0;
        for(int i = n; i >= 1; --i){
            int id = p[i].id;
            sum += RA2 - p[i].x;
            sumx[id] += (RA2 - p[i].x) * (ll)(n + 1 - i) - sum;
        }
        sort(p + 1,p + n + 1,cmpy);
        sum = 0;
        for(int i = 1; i <= n; ++i){
            int id = p[i].id;
            sum += p[i].y;
            sumy[id] = p[i].y * (ll)i - sum;
        }
        sum = 0;
        for(int i = n; i >= 1; --i){
            int id = p[i].id;
            sum += RA2 - p[i].y;
            sumy[id] += (RA2 - p[i].y) * (ll)(n + 1 - i) - sum;
        }
        sort(p + 1,p + n + 1,cmpid);
        ll ans = sumx[1] + sumy[1];
        for(int i = 2; i <= n; ++i)
            ans = min(ans,sumx[i] + sumy[i]);
        printf("%lld\n",ans >> 1);
    }
    return 0;
}