Bzoj 2834: 回家的路 dijkstra,堆优化,分层图,最短路

2834: 回家的路

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 62  Solved: 38
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Description

Input

Output

Sample Input

2 1
1 2
1 1 2 2

Sample Output

5

HINT

N<=20000,M<=100000

Source

dijkstra+堆优化+分层图

把所有的横向和纵向分开看。跑最短路即可。

注意：N这么大，不能写N^2建图。要把M个位置去建图。

 #include<bits/stdc++.h>
 using namespace std;
 #define MAXN 20010
 #define MAXM 100010
 #define INF 1e9
 struct NODE
 {
     int begin,end,value,next;
 }edge[*MAXM+];
 struct node
 {
     int x,y,id;
 }a[MAXM+];
 int cnt,Head[*MAXM+],pos[*MAXM+],Heap[*MAXM+],dis[*MAXM+],N,SIZE;
 void addedge(int bb,int ee,int vv)
 {
     edge[++cnt].begin=bb;edge[cnt].end=ee;edge[cnt].value=vv;edge[cnt].next=Head[bb];Head[bb]=cnt;
 }
 void addedge1(int bb,int ee,int vv)
 {
     addedge(bb,ee,vv);addedge(ee,bb,vv);
 }
 int read()
 {
     int s=,fh=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')fh=-;ch=getchar();}
     while(ch>=''&&ch<=''){s=s*+(ch-'');ch=getchar();}
     return s*fh;
 }
 //int xy(int x,int y){return (x-1)*n+y;}
 void Push1(int k)
 {
     int now=k,root;
     while(now>)
     {
         root=now/;
         if(dis[Heap[root]]<=dis[Heap[now]])return;
         swap(Heap[root],Heap[now]);
         swap(pos[Heap[root]],pos[Heap[now]]);
         now=root;
     }
 }
 void Insert(int k)
 {
     Heap[++SIZE]=k;pos[k]=SIZE;Push1(SIZE);
 }
 void Pop1(int k)
 {
     int now,root=k;
     pos[Heap[k]]=;Heap[k]=Heap[SIZE--];if(SIZE>)pos[Heap[k]]=k;
     while(root<=SIZE/)
     {
         now=root*;
         if(now<SIZE&&dis[Heap[now+]]<dis[Heap[now]])now++;
         if(dis[Heap[root]]<=dis[Heap[now]])return;
         swap(Heap[root],Heap[now]);
         swap(pos[Heap[root]],pos[Heap[now]]);
         root=now;
     }
 }
 void dijkstra(int start)
 {
     int i,u,v;
     for(i=;i<=N;i++)dis[i]=INF;dis[start]=;
     for(i=;i<=N;i++)Insert(i);
     while(SIZE>)
     {
         u=Heap[];Pop1(pos[u]);
         for(i=Head[u];i!=-;i=edge[i].next)
         {
             v=edge[i].end;
             if(dis[v]>dis[u]+edge[i].value){dis[v]=dis[u]+edge[i].value;Push1(pos[v]);}
         }
     }
 }
 bool cmp1(node aa,node bb)
 {
     if(aa.x==bb.x)return aa.y<bb.y;
     return aa.x<bb.x;
 }
 bool cmp2(node aa,node bb)
 {
     if(aa.y==bb.y)return aa.x<bb.x;
     return aa.y<bb.y;
 }
 int main()
 {
     int n,m,i,k,k1,bx,by,ex,ey;
     n=read();m=read();
     memset(Head,-,sizeof(Head));cnt=;
     N=*m+;
     for(i=;i<=m+;i++)a[i].x=read(),a[i].y=read(),a[i].id=i;
     sort(a+,a+m+,cmp1);
     for(i=;i<=m+;i++)
     {
         if(a[i].x==a[i+].x)addedge1(a[i].id,a[i+].id,*(a[i+].y-a[i].y));
     }
     sort(a+,a+m+,cmp2);
     for(i=;i<=m+;i++)
     {
         if(a[i].y==a[i+].y)addedge1(a[i].id+m+,a[i+].id+m+,*(a[i+].x-a[i].x));
     }
     for(i=;i<=m;i++)addedge1(i,i+m+,);
     addedge1(m+,m++m+,);addedge1(m+,m++m+,);
     dijkstra(m+);
     if(dis[m+]!=INF)printf("%d",dis[m+]);
     else printf("-1");
     /*不看n的范围的后果。。。写了个n^2的建图。。。
     for(i=1;i<=m;i++)
     {
         x=read();y=read();
         k=xy(x,y);
         addedge1(k,n*n+k,1);
     }
     for(i=1;i<=n;i++)
     {
         for(j=1;j<=n;j++)
         {
             if(i<n){k=xy(i,j);k1=xy(i+1,j);addedge1(k,k1,2);addedge1(k+n*n,k1+n*n,2);}
             if(j<n){k=xy(i,j);k1=xy(i,j+1);addedge1(k,k1,2);addedge1(k+n*n,k1+n*n,2);}
         }
     }
     N=2*n*n;
     bx=read();by=read();ex=read();ey=read();
     addedge1(xy(bx,by),xy(bx,by)+n*n,0);
     addedge1(xy(ex,ey),xy(ex,ey)+n*n,0);
     dijkstra(xy(bx,by));
     if(dis[xy(ex,ey)]!=INF)printf("%d",dis[xy(ex,ey)]);
     else printf("-1");*/
     fclose(stdin);
     fclose(stdout);
     return ;
 }