Codeforces round #260 (Div. 1) -- civilization

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• Question:
  N vertices, forests with m edges, and Q operations. Each operation: 1. Ask about the diameter of the tree where X is located. 2. Merge the trees where X and Y are located to minimize the diameter of the merged trees.
  (1? ≤?N? ≤? 3 · 105; 0? ≤?M? <?N; 1? ≤?Q? ≤? 3 · 105)
  
• Analysis:
  I didn't read the figure as forest... Tangle
  Calculate the diameter of each tree at the beginning, and then use and query the set for processing. The diameter of each merge is at least the diameter of two trees, or connect the center of the two diameters to obtain the diameter.

const int MAXN = 310000;int rt[MAXN], ans[MAXN];VI G[MAXN];bool vis[MAXN];void init(int n){    REP(i, n)    {        vis[i] = false;        ans[i] = 0;        G[i].clear();        rt[i] = i;    }}int find(int n){    return n == rt[n] ? n : rt[n] = find(rt[n]);}void merge(int a, int b){    int fa = find(a), fb = find(b);    if (fa != fb)    {        rt[fa] = fb;        ans[fb] = max(ans[fb], (ans[fb] + 1) / 2 + (ans[fa] + 1) / 2 + 1);        ans[fb] = max(ans[fa], ans[fb]);    }}int Max, id;void dfs(int u, int fa, int dep){    vis[u] = true;    REP(i, G[u].size())    {        int v = G[u][i];        if (v != fa)            dfs(v, u, dep + 1);    }    if (dep > Max)    {        Max = dep;        id = u;    }}int main(){    int n, m, q;    while (~RIII(n, m, q))    {        init(n + 1);        REP(i, m)        {            int a, b;            RII(a, b);            G[a].push_back(b);            G[b].push_back(a);            int fa = find(a), fb = find(b);            rt[fa] = fb;        }        FE(i, 1, n)        {            if (!vis[i])            {                Max = -1;                dfs(i, -1, 0);                Max = -1;                dfs(id, -1, 0);                ans[find(i)] = Max;            }        }        REP(i, q)        {            int op;            RI(op);            if (op == 2)            {                int a, b;                RII(a, b);                merge(a, b);            }            else            {                int a;                RI(a);                WI(ans[find(a)]);            }        }    }    return 0;}