[AMPPZ2014]The Captain

题面

这道题我们首先会想到优化建图；

线段树优化建图？不对不对；

那么就具体问题具体分析：

浅显的性质：有可能选择的边一定存在于按x或y排序的相邻的两个点之间；

那么O(n)建图，然后dijkstra就好了；

#include 
#define inc(i,a,b) for(register int i=a;i<=b;i++)
using namespace std;
class node{
    public:
    int x,y,id;
}points[200010];
bool cmp1(node x,node y){
    return x.x >qwq;
int dis[200010],vis[200010];
void dijkstra()
{
    qwq.push(make_pair(0,1));
    memset(dis,0x3f,sizeof(dis));
    dis[1]=0;
    while(qwq.size()){
        int u=qwq.top().second;
        qwq.pop();
        if(vis[u]) continue;
        vis[u]=1;
        for(register int i=head[u];i;i=star[i].nxt){
            int v=star[i].to;
            if(dis[v]>dis[u]+star[i].w){
                dis[v]=dis[u]+star[i].w;
                qwq.push(make_pair(-dis[v],v));
            }
        }
    }
}
int main()
{
    int n;
    scanf("%d",&n);
    inc(i,1,n) scanf("%d%d",&points[i].x,&points[i].y),points[i].id=i;
    sort(points+1,points+1+n,cmp1);
    inc(i,2,n){
        star[++cnt].add(points[i-1].id,points[i].id,points[i].x-points[i-1].x);
        star[++cnt].add(points[i].id,points[i-1].id,points[i].x-points[i-1].x);
    }
    sort(points+1,points+1+n,cmp2);
    inc(i,2,n){
        star[++cnt].add(points[i-1].id,points[i].id,points[i].y-points[i-1].y);
        star[++cnt].add(points[i].id,points[i-1].id,points[i].y-points[i-1].y);
    }
    dijkstra();
    cout<