洛谷 P3831 [SHOI2012]回家的路（最短路+分层图思想）

题意：在一个二维平面上，给你一个起点和终点，有一些车站，每次只能在车站下车，上车，或者换乘，每次换乘的时间是1，地铁不会拐弯，每次只会直线走，问你能否做地铁到达终点，如果不能打印-1

思路：有点分层图的思想，我们把横和纵的地铁先分为两层，然后直接建边，我们在第一层对第二层的连边就是换乘花费1（敲有趣的）

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int maxn=1200100;
const int maxm=1200100;
int cnt;
struct node
{
    int from,to,val,next;
}edge[maxm<<1];
int head[maxm];
LL dis[maxn];
int vis[maxn];
struct Node
{
    int val,id;
    bool operator <(const Node &b)const
    {
        if(val==b.val)return id<b.id;
        return val>b.val;
    }
};
void init()
{
    memset(head,-1,sizeof(head));
    cnt=1;
}
void addedge(int from,int to,int val)
{
    edge[cnt].from=from;
    edge[cnt].to=to;
    edge[cnt].val=val;
    edge[cnt].next=head[from];
    head[from]=cnt++;
}
LL res;
void dijkstra(int s,int e)
{
    memset(vis,0,sizeof(vis));
    memset(dis,0x7f,sizeof(dis));
    res=dis[e];
    Node now;
    priority_queue<Node>q;
    while(q.size())q.pop();
    now.val=0,now.id=s;
    dis[s]=0;
    q.push(now);
    while(!q.empty()){
        Node u=q.top();
        q.pop();
        if(vis[u.id])continue;
        vis[u.id]=1;
        for(int i=head[u.id];i!=-1;i=edge[i].next){
            int to=edge[i].to;
            if(dis[u.id]+edge[i].val<dis[to]){
                dis[to]=dis[u.id]+edge[i].val;
                Node pus;
                pus.id=to,pus.val=dis[to];
                q.push(pus);
            }
        }
    }
    return ;
}

struct point
{
    int x,y,id;
};
point p[100005];
bool cmp1(point a,point b){
    return (a.x<b.x)||(a.x==b.x&&a.y<b.y);
}
bool cmp2(point a,point b){
    return (a.y<b.y)||(a.y==b.y&&a.x<b.x);
}
int main()
{
    int n=read(),m=read();
    int s=m+1,t=m+2;
    m+=2;
    init();
    for(int i=1;i<=m;i++){
        p[i].x=read();
        p[i].y=read();
        p[i].id=i;
    }
    sort(p+1,p+1+m,cmp1);
    for(int i=1;i<m;i++){
        if(p[i].x==p[i+1].x){
            addedge(p[i].id,p[i+1].id,(p[i+1].y-p[i].y)*2);
            addedge(p[i+1].id,p[i].id,(p[i+1].y-p[i].y)*2);
        }
    }
    sort(p+1,p+1+m,cmp2);
    for(int i=1;i<m;i++){
        if(p[i].y==p[i+1].y){
            addedge(p[i].id+m,p[i+1].id+m,(p[i+1].x-p[i].x)*2);
            addedge(p[i+1].id+m,p[i].id+m,(p[i+1].x-p[i].x)*2);
        }
    }
    for(int i=1;i<=m;i++){
        addedge(p[i].id,p[i].id+m,1);
        addedge(p[i].id+m,p[i].id,1);
    }
    addedge(s,s+m,0);addedge(s+m,s,0);
    addedge(t,t+m,0);addedge(t+m,t,0);
    dijkstra(s,t);
    if(dis[t]==res)puts("-1");
    else printf("%lld
",dis[t]);
    return 0;
}