Connections between cities lca 求公共最近祖先

Connections between cities

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3795    Accepted Submission(s): 1101

Problem 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 21 3 22 4 35 2 31 44 5

 

Sample Output

Not connected6
Hint
HintHuge input, scanf recommended.
 
给一个森林，里面有好多树，给出点，判断点是否在同一个树下，如果是，求出两个点的最短距离，也就是最近公共祖先，注意起建的树，用离线Tarjan算法来求
，说实话，这个题目，到了比赛，一定不会做
#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define N 10012//node#define M 2222252//query 用来放询问的数据数组int n,m,c,h1[N],h2[N],dis[N],vis[N],fa[N],ans[M],id[N],ne,cnt;struct node{    int v,d,next;}ua[M],ub[M];void addA(int u,int v,int d){    ua[ne].v=v;    ua[ne].d=d;    ua[ne].next=h1[u];    h1[u]=ne++;}void addB(int u,int v,int d){    ub[ne].v=v;    ub[ne].d=d;    ub[ne].next=h2[u];    h2[u]=ne++;}int find(int i){    return (fa[i]==i?i:fa[i]=find(fa[i]));}void Tarjan(int u){    id[u]=cnt;    vis[u]=1;    fa[u]=u;    int i,j;    for(i=h2[u];i!=-1;i=ub[i].next)    {        int v=ub[i].v;        if(vis[v])        {            if(id[u]==id[v])            {                int rt=find(v);                ans[ub[i].d]=dis[u]+dis[v]-2*dis[rt];            }else ans[ub[i].d]=-1;        }    }    for(i=h1[u];i!=-1;i=ua[i].next)    {        int v=ua[i].v;        if(!vis[v])        {              dis[v] = dis[u] + ua[i].d;              Tarjan(v);              fa[v] = u;        }    }}int main(){    int i,j,k;    int u,v,d;    while(scanf("%d%d%d",&n,&m,&c)!=EOF)    {           for(i=1;i<=n;i++)           {               fa[i]=0;               ans[i]=id[i]=vis[i]=0;               h1[i]=h2[i]=-1;           }           ne=0;           for(i=1;i<=m;i++)           {               scanf("%d%d%d",&u,&v,&d);               addA(u,v,d);               addA(v,u,d);           }           ne=0;           for(i=1;i<=c;i++)           {               scanf("%d%d",&u,&v);               addB(u,v,i);               addB(v,u,i);           }           //printf("fhkdsjhfjkd");           for(cnt=i=1;i<=n;i++,cnt++)           {               if(!vis[i])               {   //printf("fhkdsjhfjkd");                   dis[i]=0;                   Tarjan(i);               }           }           for(i=1;i<=c;i++)           {               if(ans[i]>=0)               {                   printf("%d\n",ans[i]);               }               else               printf("Not connected\n");           }    }   return 0;}

 

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