HDU 2874 Connections between cities 在森林里面求最小距离

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Connections between cities

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3528    Accepted Submission(s): 1020

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.

 

Source

2009 Multi-University Training Contest 8 - Host by BJNU

 

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gaojie

题意是说给你一个森林，让你求两点之间的最近距离。
lca求最近公共祖先，如果不是在同一棵树上，则输出Not connected。

#include<stdio.h>#include<string.h>#define M 10007#define N 2222212int pre[M],dis[M],vis[M],ans[N],id[M];int s1[N],s2[N],t[N],d[N],p[N];int n,m,ne,cnt;int find(int x){    while(x!=pre[x])        x=pre[x];    return x;}void add(int u,int v,int w,int h[]){    t[ne]=v,d[ne]=w,p[ne]=h[u],h[u]=ne++;    t[ne]=u,d[ne]=w,p[ne]=h[v],h[v]=ne++;}void Tarjan(int u){    id[u]=cnt;    pre[u]=u;    vis[u]=1;    for(int i=s2[u];i;i=p[i])    {        int v=t[i];        if(vis[v])        {            if(id[u]==id[v])//在同一棵树下            {                int rt=find(v);//最近公共祖先                ans[d[i]]=dis[u]+dis[v]-2*dis[rt];            }            else                ans[d[i]]=-1;        }    }    for(int i=s1[u];i;i=p[i])    {        int v=t[i];        if(!vis[v])        {            dis[v]=dis[u]+d[i];            Tarjan(v);            pre[v]=u;        }    }}int main(){    //freopen("in.txt","r",stdin);    int n,m,q,i,j;    while(scanf("%d%d%d",&n,&m,&q)!=EOF)    {        for(i=1,ne=1;i<=n;i++)        {            s1[i]=s2[i]=vis[i]=ans[i]=id[i]=pre[i]=0;        }        int u,v,w;        for(i=1;i<=m;i++)        {            scanf("%d%d%d",&u,&v,&w);            add(u,v,w,s1);        }        for(i=1;i<=q;i++)        {            scanf("%d%d",&u,&v);            add(u,v,i,s2);        }        for(i=1,cnt=1;i<=n;i++,cnt++)            if(!vis[i])            {                dis[i]=0;                Tarjan(i);            }        for(i=1;i<=q;i++)            if(ans[i]>=0)                printf("%d\n",ans[i]);            else                printf("Not connected\n");    }    return 0;}