BZOJ 2834: 回家的路 Dijkstra

按照横，竖为方向跑一个最短路即可，算是水题~

#include <bits/stdc++.h>

#define N 200005

#define E 2000000

#define setIO(s) freopen(s".in","r",stdin)

using namespace std;

int n,m,tot,edges,s,T;

int hd[E],to[E],nex[E],val[E],d[E],done[E],id[E][2];

void addedge(int u,int v,int c)

{

    nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;

}

struct Point

{

    int x,y,id;

    Point(int x=0,int y=0):x(x),y(y){}

}t[N];

struct Node

{

    int u,dis;

    Node(int u=0,int dis=0): u(u),dis(dis){}

    bool operator<(Node b) const

    {

        return b.dis<dis;

    }

};

priority_queue<Node>q;

bool cmpx(Point a,Point b)

{

    return a.x==b.x?a.y<b.y:a.x<b.x;

}

bool cmpy(Point a,Point b)

{

    return a.y==b.y?a.x<b.x:a.y<b.y;

}

void Dijkstra()

{

    memset(d,0x3f,sizeof(d));

    for(d[s]=0, q.push(Node(s, 0)); !q.empty(); )

    {

        Node e = q.top(); q.pop();

        int u=e.u;

        if(done[u]) continue;

        done[u]=1;

        for(int i=hd[u];i;i=nex[i])

        {

            int v=to[i];

            if(d[v] > d[u] + val[i])

            {

                d[v] = d[u] + val[i];

                q.push(Node(v, d[v]));

            }

        }

    }

}

int main()

{

    int i,j,k;

    // setIO("input");

    scanf("%d%d",&n,&m);

    for(i=2;i<=m+1;++i)

    {

        scanf("%d%d",&t[i].x,&t[i].y);

        id[i][0]=++tot;

        id[i][1]=++tot;

        t[i].id=i;

    }

    scanf("%d%d",&t[1].x,&t[1].y);

    id[1][0]=++tot;

    id[1][1]=++tot;

    t[1].id=1;

    m+=2;

    scanf("%d%d",&t[m].x,&t[m].y);

    id[m][0]=++tot;

    id[m][1]=++tot;

    t[m].id=m;

    for(i=1;i<=m;++i)

    {

        addedge(id[i][0],id[i][1],1);

        addedge(id[i][1],id[i][0],1);

    }

    sort(t+1,t+1+m,cmpx);

    for(i=1;i<=m;i=j)

    {

        for(j=i;j<=m&&t[j].x==t[i].x;++j);

        for(k=i+1;k<j;++k)

        {

            addedge(id[t[k-1].id][1], id[t[k].id][1], 2 * (t[k].y-t[k-1].y));

            addedge(id[t[k].id][1], id[t[k-1].id][1], 2 * (t[k].y-t[k-1].y));

        }

    }

    sort(t+1,t+1+m,cmpy);

    for(i=1;i<=m;i=j)

    {

        for(j=i;j<=m&&t[j].y==t[i].y;++j);

        for(k=i+1;k<j;++k)

        {

            addedge(id[t[k-1].id][0], id[t[k].id][0], 2 * (t[k].x-t[k-1].x));

            addedge(id[t[k].id][0], id[t[k-1].id][0], 2 * (t[k].x-t[k-1].x));

        }

    }  

    s=0, T=++tot;    

    addedge(s, id[1][0], 0);

    addedge(s, id[1][1], 0);   

    addedge(id[m][0], T, 0);

    addedge(id[m][1], T, 0);

    Dijkstra();

    printf("%d\n",d[T]);

    return 0;

}

BZOJ 2834: 回家的路 Dijkstra的相关教程结束。