BZOJ 4152: [AMPPZ2014]The Captain

4152: [AMPPZ2014]The Captain

Time Limit: 20 Sec  Memory Limit: 256 MB

Description

给定平面上的n个点，定义(x1,y1)到(x2,y2)的费用为min(|x1-x2|,|y1-y2|)，求从1号点走到n号点的最小费用。

 

Input

第一行包含一个正整数n(2<=n<=200000)，表示点数。

接下来n行，每行包含两个整数x[i],y[i](0<=x[i],y[i]<=10^9)，依次表示每个点的坐标。

 

 

 

Output

一个整数，即最小费用。

 

Sample Input

5
2 2
1 1
4 5
7 1
6 7

Sample Output

2

HINT

 

Source

考虑如何减少连边。

对每一维进行分析：只需要连向它最近的一位的就可以了。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<vector>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pr;
const double pi=acos(-1);
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define Rep(i,u) for(int i=head[u];i;i=Next[i])
#define clr(a) memset(a,0,sizeof(a))
#define pb push_back
#define mp make_pair
#define fi first
#define sc second
#define pq priority_queue
#define pqb priority_queue <int, vector<int>, less<int> >
#define pqs priority_queue <int, vector<int>, greater<int> >
#define vec vector
ld eps=1e-9;
ll pp=1000000007;
ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;}
ll powmod(ll a,ll b,ll pp){ll ans=1;for(;b;b>>=1,a=mo(a*a,pp))if(b&1)ans=mo(ans*a,pp);return ans;}
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
//void add(int x,int y,int z){ v[++e]=y; next[e]=head[x]; head[x]=e; cost[e]=z; }
int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1};
ll read(){ ll ans=0; char last=' ',ch=getchar();
while(ch<'0' || ch>'9')last=ch,ch=getchar();
while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
if(last=='-')ans=-ans; return ans;
}
const int M=800005,N=200005;
int v[M],Next[M],head[N],vis[N],n,e,X[N],Y[N];
ll dis[N],cost[M];
struct nodeQ{
    ll v; int i; 
    friend bool operator <(nodeQ a,nodeQ b){
        return a.v>b.v;
    }
};
struct node{
    int x,y,i;
}f[N];
bool cmpx(node a,node b){
    return a.x<b.x;
}
bool cmpy(node a,node b){
    return a.y<b.y;
}
#include<queue>
priority_queue<nodeQ> Q;
void add(int x,int y){v[++e]=y; Next[e]=head[x]; head[x]=e; cost[e]=min(abs(X[x]-X[y]),abs(Y[x]-Y[y]));  }
void Dij(){
    for (int i=1;i<=n;i++) dis[i]=1e17,vis[i]=0;
    dis[1]=0;
    Q.push((nodeQ){dis[1],1});
    while (!Q.empty()){
        int u=Q.top().i; Q.pop(); if (vis[u]) continue;
        vis[u]=1; 
        for (int i=head[u];i;i=Next[i]){
            if (dis[v[i]]>dis[u]+cost[i]){
                dis[v[i]]=dis[u]+cost[i];
                Q.push((nodeQ){dis[v[i]],v[i]});
            }
        }
    }
}
int main(){
    n=read();
    for (int i=1;i<=n;i++) f[i].x=X[i]=read(),f[i].y=Y[i]=read(),f[i].i=i;
    sort(f+1,f+n+1,cmpx);
    for (int i=1;i<n;i++) add(f[i].i,f[i+1].i),add(f[i+1].i,f[i].i);;
    sort(f+1,f+n+1,cmpy);
    for (int i=1;i<n;i++) add(f[i].i,f[i+1].i),add(f[i+1].i,f[i].i);
    Dij();
    printf("%lld",dis[n]);
    return 0;
}