HDU2874【倍增、ST】

题目链接【https://vjudge.net/problem/HDU-2874】

题意： 输入一个森林，总节点不超过N(N<10000),由C次询问（C<1000000）,每次询问两个点，如果来联通输出，两点之间的距离，如果不来联通，输出“Not connected”;

思路：首先判断u,v两个点在不在同一个树上，用并查集来做，如果在，就求两者的LCA，输入距离dis[u->v]=dis[u]+dis[v]-2*dis[LCA(u,v)]。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 10050;
int N, M, C;
struct node
{
    int id, next, len;
} E[maxn << 1];
int head[maxn << 1], num;
void initlist()
{
    memset(head, -1, sizeof(head));
    num = 0;
}
void adde(int u, int v, int len)
{
    E[num].id = u;
    E[num].len = len;
    E[num].next = head[v];
    head[v] = num++;
}
//----------------------------------------邻接表
int fa[maxn<<1];
void initfa()
{
    for(int i = 0; i <= N; i++)
        fa[i] = i;
}
int Find(int id)
{
    if(id == fa[id]) return id;
    else return fa[id] = Find(fa[id]);
}
void addu(int u, int v)
{
    int x = Find(u);
    int y = Find(v);
    if(x != y)    fa[x] = y;
}
//----------------------------------------并查集
int p[16][maxn], dep[maxn], dis[maxn], vis[maxn];
void DFS(int u, int FA)
{
    vis[u] = 1;
    for(int l = head[u]; l != -1; l = E[l].next)
    {
        int id = E[l].id;
        if(id == FA||vis[id]) continue;
        dep[id] = dep[u] + 1;
        dis[id] = dis[u] + E[l].len;
        p[0][id] = u;
        DFS(id, u);
    }
}
void initlca()
{
    memset(p, -1, sizeof(p));
    memset(dep,0,sizeof(dep));
    memset(dis, 0, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= N; i++)
    {
        if(!vis[i])
            DFS(i, -1);
    }
    for(int k = 0; k + 1 <= 15; k++)
        for(int u = 1; u <= N; u++)
        {
            if(p[k][u] < 0) p[k + 1][u] = -1;
            else
                p[k + 1][u] = p[k][p[k][u]];
        }
}
int LCA(int u, int v)
{
    if(dep[u] > dep[v]) swap(u, v);
    for(int k = 0; k <= 15; k++)
    {
        if((dep[v] - dep[u]) >> k & 1)
            v = p[k][v];
    }
    if(u == v) return u;
    for(int k = 15; k >= 0; k--)
    {
        if(p[k][u] != p[k][v])
        {
            u = p[k][u];
            v = p[k][v];
        }
    }
    return p[0][u];
}
//----------------------------------------LCA
int main ()
{
    while(~scanf("%d%d%d", &N, &M, &C))
    {
        initlist();
        initfa();
        int u, v, ds;
        for(int i = 1; i <= M; i++)
        {
            scanf("%d%d%d", &u, &v, &ds);
            adde(u, v, ds);
            adde(v, u, ds);
            addu(u, v);
        }
        initlca();
        for(int i=1;i<=C;i++)
        {
            scanf("%d%d",&u,&v);
            int x=Find(u);
            int y=Find(v);
            if(x==y)
            {
                printf("%d
",dis[u]+dis[v]-2*dis[LCA(u,v)]);
            }
            else
                printf("Not connected
");
        }
    }
    return 0;
}

//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 100050;
//---------------------------------//邻接表
struct node
{
    int to, len, next;
} E[MAXN * 2];
int p[MAXN * 2], tot;
void init()
{
    memset(p, -1, sizeof(p));
    tot = 0;
}
void add(int u, int v, int len)
{
    E[tot].to = v;
    E[tot].len = len;
    E[tot].next = p[u];
    p[u] = tot++;
}
//---------------------------------//并查集
int fa[MAXN * 2];
void init2(int n)
{
    for(int i = 1; i <= n; i++)
        fa[i] = i;
}
int Find(int x)
{
    if(x == fa[x])
        return fa[x];
    else
        return fa[x] = Find(fa[x]);
}
void join (int i, int j)
{
    int x = Find(i);
    int y = Find(j);
    if(x != y) fa[y] = x;
}
//----------------------------------//LAC
int ver[MAXN * 2], vis[MAXN], fst[MAXN], R[MAXN * 2];
int dis[MAXN], KT[MAXN * 2], dp[MAXN * 2][25];
void init3()
{
    memset(vis, 0, sizeof(vis));
    tot = 0;
}
void DFS(int u, int dep, int len)
{
    vis[u] = 1;
    ver[++tot] = u;
    R[tot] = dep;
    fst[u] = tot;
    dis[u] = len;
    for(int k = p[u]; k != -1; k = E[k].next)
    {
        int v = E[k].to;
        if(vis[v]) continue;
        DFS(v, dep + 1, len + E[k].len);
        ver[++tot] = u;
        R[tot] = dep;
    }
}
void ST(int n)
{
    KT[1] = 0;
    for(int i = 2; i <= n; i++)
        KT[i] = KT[i / 2] + 1;
    for(int i = 1; i <= n; i++)
        dp[i][0] = i;
    for(int j = 1; j <= KT[n]; j++)
        for(int i = 1; i + (1 << j) - 1 <= n; i++)
        {
            int x = dp[i][j - 1];
            int y = dp[i + (1 << (j - 1))][j - 1];
            dp[i][j] = R[x] < R[y] ? x : y;
        }
}
int RMQ(int l, int r)
{
    int k = KT[r - l + 1];
    int x = dp[l][k];
    int y = dp[r - (1 << k) + 1][k];
    return R[x] < R[y] ? x : y;
}
int LCA(int u, int v)
{
    int l = fst[u], r = fst[v];
    if(l > r) swap(l, r);
    int t = RMQ(l, r);
    return ver[t];
}
//-------------------------------------//主函数
int N, M, C;
int main ()
{
    while(~scanf("%d%d%d", &N, &M, &C))
    {
        init();//
        init2(N);
        int u, v, len;
        for(int i = 1; i <= M; i++)
        {
            scanf("%d%d%d", &u, &v, &len);
            add(u, v, len);
            add(v, u, len);
            join(u, v);
        }
        init3();
        for(int i = 1; i <= N; i++)
        {
            if(vis[i]) continue;
            DFS(i, 1, 0);
        }
        ST(2 * N - 1);
        for(int i = 1; i <= C; i++)
        {
            scanf("%d%d", &u, &v);
            int x = Find(u);
            int y = Find(v);
            if(x != y)
            {
                printf("Not connected
");
                continue;
            }
            int lca = LCA(u, v);
            printf("%d
", dis[u] + dis[v] - 2 * dis[lca]);
        }
    }
    return 0;
}