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。

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<cstdlib>
 7 #include<vector>
 8 using namespace std;
 9 typedef long long ll;
10 typedef long double ld;
11 typedef pair<int,int> pr;
12 const double pi=acos(-1);
13 #define rep(i,a,n) for(int i=a;i<=n;i++)
14 #define per(i,n,a) for(int i=n;i>=a;i--)
15 #define Rep(i,u) for(int i=head[u];i;i=Next[i])
16 #define clr(a) memset(a,0,sizeof(a))
17 #define pb push_back
18 #define mp make_pair
19 #define fi first
20 #define sc second
21 #define pq priority_queue
22 #define pqb priority_queue <int, vector<int>, less<int> >
23 #define pqs priority_queue <int, vector<int>, greater<int> >
24 #define vec vector
25 ld eps=1e-9;
26 ll pp=1000000007;
27 ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;}
28 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;}
29 void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
30 //void add(int x,int y,int z){ v[++e]=y; next[e]=head[x]; head[x]=e; cost[e]=z; }
31 int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1};
32 ll read(){ ll ans=0; char last=' ',ch=getchar();
33 while(ch<'0' || ch>'9')last=ch,ch=getchar();
34 while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
35 if(last=='-')ans=-ans; return ans;
36 }
37 const int M=800005,N=200005;
38 int v[M],Next[M],head[N],vis[N],n,e,X[N],Y[N];
39 ll dis[N],cost[M];
40 struct nodeQ{
41     ll v; int i; 
42     friend bool operator <(nodeQ a,nodeQ b){
43         return a.v>b.v;
44     }
45 };
46 struct node{
47     int x,y,i;
48 }f[N];
49 bool cmpx(node a,node b){
50     return a.x<b.x;
51 }
52 bool cmpy(node a,node b){
53     return a.y<b.y;
54 }
55 #include<queue>
56 priority_queue<nodeQ> Q;
57 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]));  }
58 void Dij(){
59     for (int i=1;i<=n;i++) dis[i]=1e17,vis[i]=0;
60     dis[1]=0;
61     Q.push((nodeQ){dis[1],1});
62     while (!Q.empty()){
63         int u=Q.top().i; Q.pop(); if (vis[u]) continue;
64         vis[u]=1; 
65         for (int i=head[u];i;i=Next[i]){
66             if (dis[v[i]]>dis[u]+cost[i]){
67                 dis[v[i]]=dis[u]+cost[i];
68                 Q.push((nodeQ){dis[v[i]],v[i]});
69             }
70         }
71     }
72 }
73 int main(){
74     n=read();
75     for (int i=1;i<=n;i++) f[i].x=X[i]=read(),f[i].y=Y[i]=read(),f[i].i=i;
76     sort(f+1,f+n+1,cmpx);
77     for (int i=1;i<n;i++) add(f[i].i,f[i+1].i),add(f[i+1].i,f[i].i);;
78     sort(f+1,f+n+1,cmpy);
79     for (int i=1;i<n;i++) add(f[i].i,f[i+1].i),add(f[i+1].i,f[i].i);
80     Dij();
81     printf("%lld",dis[n]);
82     return 0;
83 }
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posted @ 2017-10-22 20:47  SXia  阅读(174)  评论(0编辑  收藏  举报