POJ 2874 LCA 树上任意两点距离

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本题说了是无环图，所以就是一片森林了。而对于树上的任意两点，我们可以用LCA求其距离。距离为两个子节点到根的距离和减去最近祖先到根的距离的2倍。具体画图便可看出来。

并且图是无向图，所以LCA时需要进行标记

POJ 1986同这道题基本一样

/*ID: CUGB-wwjPROG:LANG: C++*/#include <iostream>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cctype>#include <string>#include <cstring>#include <cmath>#include <ctime>#define INF 1111111111#define MAXN 10005#define MAXM 1000005#define L(x) x<<1#define R(x) x<<1|1#define eps 1e-4using namespace std;struct node{    int v, w;    node(){}    node(int a, int b){ v = a; w = b;}};vector<node>tree[MAXN], ask[MAXN];  //tree里w代表权值，ask里w代表序号int dis[MAXN], father[MAXN], vis[MAXN], ans[MAXM];int n, m, c;void init(){    for(int i = 0; i <= n; i++)    {        dis[i] = 0;        father[i] = i;        vis[i] = 0;        tree[i].clear();        ask[i].clear();    }}int find(int x){    if(father[x] == x) return x;    int t = find(father[x]);    father[x] = t;    return t;}void LCA(int u, int w, int root){    dis[u] = w;    father[u] = u;    vis[u] = root;    int size = tree[u].size();    for(int i = 0; i < size; i++)    {        if(!vis[tree[u][i].v])        {            LCA(tree[u][i].v, tree[u][i].w + w, root);            father[tree[u][i].v] = u;        }    }    size = ask[u].size();    for(int i = 0; i < size; i++)    {        if(vis[ask[u][i].v])        {            if(vis[ask[u][i].v] == root)            ans[ask[u][i].w] = dis[u] + dis[ask[u][i].v] - 2 * dis[find(ask[u][i].v)];            else ans[ask[u][i].w] = -1;        }    }}int main(){    int u, v, w;    while(scanf("%d%d%d", &n, &m, &c) != EOF)    {        init();        for(int i = 1; i <= m; i++)        {            scanf("%d%d%d", &u, &v, &w);            node tmp1(v, w);            node tmp2(u, w);            tree[u].push_back(tmp1);            tree[v].push_back(tmp2);        }        for(int i = 1; i <= c; i++)        {            scanf("%d%d", &u, &v);            node tmp1(v, i);            node tmp2(u, i);            ask[u].push_back(tmp1);            ask[v].push_back(tmp2);        }        for(int i = 1; i <= n; i++)        {            if(!vis[i]) LCA(i, 0, i);        }        for(int i = 1; i <= c; i++)        {            if(ans[i] == -1) printf("Not connected\n");            else printf("%d\n", ans[i]);        }    }    return 0;}