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=1;i<=n;i++)
    scanf("%d%d",&v[i].a,&v[i].b),v[i].id=i;
    sort(v+1,v+1+n,cmp1);
    for (i=1;i<n;i++)
    {
        link(v[i].id,v[i+1].id,abs(v[i].a-v[i+1].a));
        link(v[i+1].id,v[i].id,abs(v[i].a-v[i+1].a));
    }
    sort(v+1,v+1+n,cmp2);
    for (i=1;i<n;i++)
    {
        link(v[i].id,v[i+1].id,abs(v[i].b-v[i+1].b));
        link(v[i+1].id,v[i].id,abs(v[i].b-v[i+1].b));
    }
    int inf=1000000000;
    for (i=2;i<=n;i++) dis[i]=inf;
    Q.push(mp(0,1));
    
    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
",dis[n]);
}