HDU 2874 Connections between cities （LCA離線&&線上RMQ，4級）__RMQ&LCA

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

HintHuge input, scanf recommended. 

 思路：LCA離線

#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線上轉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)//一遍歐拉路徑{ int v;  to[dfs_clock]=u;//存歐拉路徑  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)///虛點 0 ，虛邊得有值，不然查到0不一定是根點    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);    }  }}