bzoj4152 [AMPPZ2014]The Captain

　　最短路，先将x排序，然后把排序后权值相邻的点连边，再把y排序，也把权值相邻的点连边，求一遍1到n的最短路就好啦。

　　代码

 #include<cstdio>
 #include<queue>
 #include<algorithm>
 #define mp make_pair
 #define fi first
 #define sc second
 #define N 1000000
 using namespace std;
 typedef long long ll;
 typedef pair<int,int> P;
 int n,i,dp,p[N],pre[N],tt[N],ww[N],flag[N];
 int dis[N];
 priority_queue<P,vector<P>,greater<P> > Q;
 struct g{
     int a,b,id;
 }v[N];
 bool cmp1(g u,g v)
 {
     return u.a<v.a;
 }
 bool cmp2(g u,g v)
 {
     return u.b<v.b;
 }
 
 void link(int x,int y,int z)
 {
     dp++;pre[dp]=p[x];p[x]=dp;tt[dp]=y;ww[dp]=z;
 }
 long long ans,f[N],sum[N];
 int main()
 {
     scanf("%d",&n);
     for (i=;i<=n;i++)
     scanf("%d%d",&v[i].a,&v[i].b),v[i].id=i;
     sort(v+,v++n,cmp1);
     for (i=;i<n;i++)
     {
         link(v[i].id,v[i+].id,abs(v[i].a-v[i+].a));
         link(v[i+].id,v[i].id,abs(v[i].a-v[i+].a));
     }
     sort(v+,v++n,cmp2);
     for (i=;i<n;i++)
     {
         link(v[i].id,v[i+].id,abs(v[i].b-v[i+].b));
         link(v[i+].id,v[i].id,abs(v[i].b-v[i+].b));
     }
     int inf=;
     for (i=;i<=n;i++) dis[i]=inf;
     Q.push(mp(,));
 
     while (!Q.empty())
     {
         P x=Q.top();Q.pop();
         if (x.fi!=dis[x.sc]) continue;
         i=p[x.sc];
         while (i)
         {
             if (dis[x.sc]+ww[i]<dis[tt[i]])
             {
                 dis[tt[i]]=dis[x.sc]+ww[i];
                 Q.push(mp(dis[tt[i]],tt[i]));
             }
             i=pre[i];
         }
     }
     printf("%d\n",dis[n]);
 }