How far away ？【HDU - 2586】【LCA - Tarjan离线】

HDU - 2586

• 题意：给一颗无根树，求两个结点间的最短路径长度。

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 思路：

因为是在树上求最短路，那么肯定是通过LCA的那条路径为最短。

那么我们应该怎么求路径长度呢？

我们以任意结点为根，将无根树转化为有根树，然后dis[ i ]表示结点 i 到根结点的最短距离，所以dis<u, v> = dis[ u ] + dis[ v ] - dis[ LCA(u, v) ] * 2

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AC CODE 

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;

inline int read()
{
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -f; c = getchar(); }
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

const int maxN = 40004;
int n, m;
int dis[maxN];

struct EDGE{
    int adj, to, w;
    EDGE(int a = -1, int b = 0, int c = 0): adj(a), to(b), w(c) {}
}edge[maxN << 1];
int head[maxN], cnt;

void add_edge(int u, int v, int w)
{
    edge[cnt] = EDGE(head[u], v, w);
    head[u] = cnt ++;
}

int root[maxN];
int Find(int x) { return root[x] == x ? x : root[x] = Find(root[x]); }
bool Same(int x, int y) { return Find(x) == Find(y); }
void Merge(int x, int y) { root[Find(y)] = Find(x); }

bool vis[maxN];
int lca[maxN];
vector<pair<int, int > > ques[maxN];
vector<pair<int, int > >info;

void Tarjan(int u, int fa)
{
    vis[u] = true;
    for(int i = head[u]; ~i; i = edge[i].adj)
    {
        int v = edge[i].to;
        if(v != fa)
        {
            dis[v] = dis[u] + edge[i].w;
            Tarjan(v, u);
            if(!Same(u, v)) Merge(u, v);
        }
    }
    if(!ques[u].empty())
    {
        int siz = ques[u].size();
        for(int i = 0; i < siz; ++ i )
        {
            int v = ques[u][i].first;
            if(vis[v])
                lca[ques[u][i].second] = Find(v);
        }
    }
}

void init()
{
    for(int i = 0; i <= n; ++ i )
    {
        head[i] = -1;
        root[i] = i;
        vis[i] = false;
        dis[i] = 0;
        ques[i].clear();
    }
    info.clear();
    dis[0] = dis[1] = 0;
    cnt = 0;
}

int main()
{
    int TAT = read();
    while(TAT -- )
    {
        n = read(); m = read();
        init();
        for(int i = 0; i < n - 1; ++ i )
        {
            int u, v, w;
            u = read(); v = read(); w = read();
            add_edge(u, v, w);
            add_edge(v, u, w);
        }
        for(int i = 0; i < m; ++ i )
        {
            int u, v;
            u = read(); v = read();
            ques[u].push_back(make_pair(v, i));
            ques[v].push_back(make_pair(u, i));
            info.push_back(make_pair(u, v));
        }
        Tarjan(1, 0);
        for(int i = 0; i < m; ++ i )
            printf("%d\n", dis[info[i].first] + dis[info[i].second] - dis[lca[i]] * 2);
    }
    return 0;
}