HDU 2874 Connections between cities LCA水题 注意是森林，不是树

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 2

1 3 2

2 4 3

5 2 3

1 4

4 5

 

 

Sample Output

Not connected

6

 

 

 

 

 

注意：这道题不一定是树，有可能是森林，

只要学树链剖分，加一个top[i] 表示节点i的根是top[i] 就好啦。

 

 

length=siz[u]+siz[v]-2*siz[lca(u,v)]

 

 

 

 

#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

#define LL long long

const int maxn=10000+5;

struct edge
{
    int to,w,next;
}edge[maxn<<1];
int head[maxn];
int tot;

int siz[maxn];
int dep[maxn];
int p[maxn][25];
int top[maxn];

void init(int n)
{
    tot=0;
    memset(head,-1,sizeof(head));
    memset(siz,-1,sizeof(siz));
    memset(dep,0,sizeof(dep));

    for(int i=1;i<=n;i++)
        top[i]=i;

    for(int i=1;i<=n;i++)
        for(int j=0;j<25;j++)
            p[i][j]=-1;
}

void addedge(int u,int v,int w)
{
    edge[tot].to=v;
    edge[tot].w=w;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void dfs(int u)
{
    for(int i=head[u];~i;i=edge[i].next)
    {
        int v=edge[i].to;
        int w=edge[i].w;
        if(!dep[v])
        {
            siz[v]=siz[u]+w;
            dep[v]=dep[u]+1;
            p[v][0]=u;
            top[v]=top[u];
            dfs(v);
        }
    }
}

void init_lca(int n)
{
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i<=n;i++)
        {
            if(p[i][j-1]!=-1)
            {
                p[i][j]=p[p[i][j-1]][j-1];
            }
        }
    }
}

LL solve(int n,int u,int v)
{
    if(dep[u]<dep[v])
        swap(u,v);

    int init_u=u;
    int init_v=v;

    int cnt;
    for(cnt=0;(1<<cnt)<=dep[u];cnt++)
        ;
    cnt--;

    for(int j=cnt;j>=0;j--)
    {
        if(dep[u]-(1<<j)>=dep[v])
            u=p[u][j];
    }
    if(u==v)
        return (LL)(siz[init_u]-siz[v]);
    else
    {
        for(int j=cnt;j>=0;j--)
        {
            if(p[u][j]!=-1&&p[u][j]!=p[v][j])
            {
                u=p[u][j];
                v=p[v][j];
            }
        }
        return (LL)(siz[init_u]+siz[init_v]-2*siz[p[u][0]]);
    }
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int m,c;
        scanf("%d%d",&m,&c);

        init(n);

        for(int i=1;i<=m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
        }

        for(int i=1;i<=n;i++)
        {
            if(siz[i]==-1)
            {
                siz[i]=0;
                dfs(i);
            }
        }

        init_lca(n);

        for(int i=0;i<c;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);

            if(top[u]!=top[v])
            {
                printf("Not connected\n");
            }
            else
            {
                printf("%lld\n",solve(n,u,v));
            }
        }

    }

    return 0;
}