HDU 4311 树状数组+二分

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题意：给出10W个点，找出一个点使得每点按网格走到它的步数和最小，求最小步数和。每步这么走Eg: (x,y) can be reached from (x-1,y), (x+1,y), (x, y-1), (x, y+1).。

a点到达b点的步数为abs(b.x-a.x)+abs(b.y-a.y) 那么a点到所有点的步数和为abs(pi.x-a.x)+abs(pi.y-ay)所以按照从小到大的顺序吧x，y的值排序，二分查找a.x a.y在数组中的位置，就是给上面那个求和在打开就变成abs(numx*ax-sumpi.x(1~numx))+abs(pi.x(numx+1~n)-(n-numx)*a.x)就是x的和同理求出y的和就可以了意思就是这样

ans=(x1+x2+x3+...+xn)-n*xi+(y1+y2+...+yn)-n*yi; 不过需要考虑大小关系

(括号里面的sum部分可以用树状数组、线段树来维护)

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define maxn 100005long long treex[maxn<<1],treey[maxn<<1],datax[maxn],datay[maxn],sumx,sumy;struct cor{    long long x,y;} data[maxn];int n;inline int Lowbit(int x){    return x&(-x);}inline void Update(int pos,long long val,long long *tree){    while(pos<=maxn)        tree[pos]+=val,pos+=Lowbit(pos);}inline long long Query(int x,long long *tree){    long long sum=0;    while(x>0)        sum+=tree[x],x-=Lowbit(x);    return sum;}inline int find(long long val,long long *data){    int l=0,r=n-1,mid;    while(l<=r)    {        mid=(l+r)>>1;        if(val==data[mid])            return mid+1;        else if(val<data[mid])            r=mid-1;        else            l=mid+1;    }}long long ab(long long a){    return a<0?-a:a;}long long getans(cor a){    long long sum,nowx,nowy;    int numx,numy;    numx=find(a.x,datax),numy=find(a.y,datay);    nowx=Query(numx,treex),nowy=Query(numy,treey);    sum=ab((long long)numx*a.x-nowx)+ab(sumx-nowx-(long long)(n-numx)*a.x)+ab((long long)numy*a.y-nowy)+ab(sumy-nowy-(long long)(n-numy)*a.y);    return sum;}int main(){    int t;    long long ans;    scanf("%d",&t);    while(t--)    {        sumx=sumy=0;        memset(treex,0,sizeof(treex));        memset(treey,0,sizeof(treey));        scanf("%d",&n);        for(int i=0; i<n; i++)            scanf("%I64d%I64d",&data[i].x,&data[i].y),                datax[i]=data[i].x,datay[i]=data[i].y,                    sumx+=data[i].x,sumy+=data[i].y;        sort(datax,datax+n);        sort(datay,datay+n);        for(int i=0; i<n; i++)            Update(i+1,datax[i],treex),Update(i+1,datay[i],treey);        ans=1e17;        for(int i=0; i<n; i++)            ans=min(ans,getans(data[i]));        printf("%I64d\n",ans);    }    return 0;}