[hdu4311]Meeting point-1
题意：在整数坐标轴上找一个距离所有给定点距离最小的点。
解题关键：对x和y分别处理，前缀和预处理所有点到最小点的距离，每点的$sum$等于左边的贡献+右边的贡献，最后取$min$即可。
复杂度：$O(nlogn)$
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cstdlib> 5 #include<iostream> 6 #include<cmath> 7 using namespace std; 8 typedef long long ll; 9 const ll INF = 1ll<<61-1; 10 struct node{ 11 ll x,y,sum; 12 }p[100002]; 13 bool cmp1(const node &a,const node &b){ 14 return a.x<b.x; 15 } 16 17 bool cmp2(const node &a,const node &b){ 18 return a.y<b.y; 19 } 20 ll dis[100002]; 21 ll sum1[100002]; 22 int main(){ 23 ll t; 24 scanf("%I64d",&t); 25 while(t--){ 26 memset(p,0,sizeof p); 27 ll n; 28 scanf("%I64d",&n); 29 for(int i=1;i<=n;i++) scanf("%I64d%I64d",&p[i].x,&p[i].y); 30 sort(p+1,p+n+1,cmp1); 31 for(int i=1;i<=n;i++) dis[i]=p[i].x-p[1].x; 32 for(int i = 1;i <= n;i++) sum1[i]=sum1[i-1]+dis[i]; 33 for(int i=1;i<=n;i++){ 34 p[i].sum+=abs((i-1)*dis[i]-sum1[i-1]); 35 p[i].sum+=abs((sum1[n]-sum1[i])-dis[i]*(n-i)); 36 } 37 sort(p+1,p+n+1,cmp2); 38 for(int i=1;i<=n;i++) dis[i]=p[i].y-p[1].y; 39 for(int i = 1;i <= n;i++) sum1[i]=sum1[i-1]+dis[i]; 40 for(int i=1;i<=n;i++){ 41 p[i].sum+=abs((i-1)*dis[i]-sum1[i-1]); 42 p[i].sum+=abs((sum1[n]-sum1[i])-dis[i]*(n-i)); 43 } 44 ll ma=INF; 45 for(int i=1;i<=n;i++){ 46 if(p[i].sum<ma){ 47 ma=p[i].sum; 48 } 49 } 50 printf("%lld\n",ma); 51 } 52 53 }
