HDU 2874 Connections between cities （LCA离线&&在线RMQ，4级）

E - Connections between cities

Crawling in process... Crawling failed Time Limit:5000MS    Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

Appoint description: System Crawler (2013-05-30)

Description

After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city. For most of roads had been totally destroyed during the war, there might be no path between two cities, no circle exists as well.
Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.

 

Input

Input consists of multiple problem instances.For each instance, first line contains three integers n, m and c, 2<=n<=10000, 0<=m<10000, 1<=c<=1000000. n represents the number of cities numbered from 1 to n. Following m lines, each line has three integers i, j and k, represent a road between city i and city j, with length k. Last c lines, two integers i, j each line, indicates a query of city i and city j.

 

Output

For each problem instance, one line for each query. If no path between two cities, output “Not connected”, otherwise output the length of the shortest path between them.

 

Sample Input

      
      
      
      

       
       
       
       5 3 2
1 3 2
2 4 3
5 2 3
1 4
4 5
      
      
      
      

 

Sample Output

      
      
      
      

       
       
       
       Not connected
6


      
      
      
      

Hint

Hint

Huge input, scanf recommended.
 

 思路：LCA离线

#include<iostream>
#include<cstring>
#include<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int nn=1e4+9;
const int mm=2e6+nn+nn;
class Edge
{
  public:int v,next,w;
}e[mm];
int head[nn],qhead[nn],edge,ans[mm],dis[nn];
bool vis[nn];
int rt[nn],N,M,C,id[nn];
void data()
{
  clr(head,-1);clr(qhead,-1);edge=0;
}
void add(int u,int v,int w,int*h)
{
 e[edge].v=v;e[edge].w=w;e[edge].next=h[u];h[u]=edge++;
}
int find(int x)
{
  if(rt[x]^x)
    rt[x]=find(rt[x]);
  return rt[x];
}
void tarjan(int u,int bcc)
{ int v;
  id[u]=bcc;
  vis[u]=1;rt[u]=u;
  for(int i=head[u];~i;i=e[i].next)
  {
    v=e[i].v;
    if(vis[v])continue;
    dis[v]=dis[u]+e[i].w;
    tarjan(v,bcc);rt[v]=u;
  }
  for(int i=qhead[u];~i;i=e[i].next)
  {
    v=e[i].v;
    if(!vis[v])continue;
    ans[e[i].w]=(id[u]==id[v])?dis[u]+dis[v]-dis[find(v)]*2:-1;
  }
}
void getans()
{ clr(vis,0);clr(id,0);
  int bcc=0;
  FOR(i,1,N)
  if(!vis[i])
  dis[i]=0,tarjan(i,++bcc);
}
int main()
{ int a,b,c;
  while(~scanf("%d%d%d",&N,&M,&C))
  {
    data();
    FOR(i,1,M)
    {
      scanf("%d%d%d",&a,&b,&c);
      add(a,b,c,head);add(b,a,c,head);
    }
    FOR(i,1,C)
    {
      scanf("%d%d",&a,&b);add(a,b,i,qhead);add(b,a,i,qhead);
    }
    getans();
    FOR(i,1,C)
    if(ans[i]<0)printf("Not connected\n");
    else printf("%d\n",ans[i]);
  }
}

LCA在线转RMQ

#include<iostream>
#include<cstring>
#include<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
#define ll(x) (1<<x)
using namespace std;
const int nn=4e4+9;
const int mm=4e4+9;
class Edge
{
  public:int v,next,w;
}e[mm];
int head[nn],edge,dis[nn],to[nn],dfs_clock;
int vis[nn],rmq[nn][30];
int rt[nn],N,M,C,bit[nn];
void data()
{
  clr(head,-1);edge=0;
}
void add(int u,int v,int w,int*h)
{
 e[edge].v=v;e[edge].w=w;e[edge].next=h[u];h[u]=edge++;
}
int find(int x)
{
  if(rt[x]^x)
    rt[x]=find(rt[x]);
  return rt[x];
}
void uni(int a,int b)
{
  a=find(a);b=find(b);
  rt[a]=b;
}
void dfs(int u,int dep)//一遍欧拉路径
{ int v;
  to[dfs_clock]=u;//存欧拉路径
  dis[u]=dep;
  vis[u]=dfs_clock++;
  for(int i=head[u];~i;i=e[i].next)
  {
    v=e[i].v;
    if(vis[v]==-1)
    {
      dfs(v,dep+e[i].w);
      to[dfs_clock++]=u;
    }
  }
}
void ST(int N)
{
  bit[0]=-1;
  FOR(i,1,N)bit[i]=(i&(i-1))==0?bit[i-1]+1:bit[i-1];
  FOR(i,0,N)
  rmq[i][0]=dis[ to[i] ];
  FOR(i,1,bit[N])
  for(int j=0;j+ll(i)-1<=N;++j)
    rmq[j][i]=min(rmq[j][i-1],rmq[j+ll(i-1)][i-1]);
}
int RMQ(int l,int r)
{
  int t=bit[r-l+1];
  r-=ll(t)-1;
  return min(rmq[l][t],rmq[r][t]);
}
int main()
{ int a,b,c;
  while(~scanf("%d%d%d",&N,&M,&C))
  {
    data();clr(vis,-1);
    FOR(i,0,N)rt[i]=i;
    FOR(i,1,M)
    {
      scanf("%d%d%d",&a,&b,&c);
      add(a,b,c,head);add(b,a,c,head);
      uni(a,b);
    }
    FOR(i,1,N)///虚点 0 ，虚边得有值，不然查到0不一定是根点
    if(rt[i]==i)
    add(0,i,1,head),add(i,0,1,head);
    dfs_clock=0;
    dfs(0,0);
    ST(dfs_clock-1);
    FOR(i,1,C)
    {
      scanf("%d%d",&a,&b);
      int ta=vis[a];
      int tb=vis[b];
      if(ta>tb)swap(ta,tb);
      int ddd=RMQ(ta,tb);
      if(ddd==0)printf("Not connected\n");
      else printf("%d\n",dis[a]+dis[b]-2*ddd);
    }
  }
}