codeforces 455C 并查集

传送门

给n个点， 初始有m条边， q个操作。

每个操作有两种， 1是询问点x所在的连通块内的最长路径， 就是树的直径。 2是将x, y所在的两个连通块连接起来，并且要合并之后的树的直径最小，如果属于同一个连通块就忽视这个操作。

先dfs出每个连通块的初始直径， 然后合并的话就是len[x] = max( (len[x]+1)/2+(len[y]+1)/2+1, max(len[x], len[y]));  然后搞一搞就好了。

一开始写bfs写挫了一直超时，  只好改成dfs......

 #include<bits/stdc++.h>
 using namespace std;
 #define pb(x) push_back(x)
 #define ll long long
 #define mk(x, y) make_pair(x, y)
 #define lson l, m, rt<<1
 #define mem(a) memset(a, 0, sizeof(a))
 #define rson m+1, r, rt<<1|1
 #define mem1(a) memset(a, -1, sizeof(a))
 #define mem2(a) memset(a, 0x3f, sizeof(a))
 #define rep(i, a, n) for(int i = a; i<n; i++)
 #define ull unsigned long long
 typedef pair<int, int> pll;
 const double PI = acos(-1.0);
 const double eps = 1e-;
 const int mod = 1e9+;
 const int inf = ;
 const int dir[][] = { {-, }, {, }, {, -}, {, } };
 const int maxn = 3e5+;
 int f[maxn], num[maxn], head[maxn*], dis[maxn], cnt, q[maxn*];
 struct node
 {
     int to, nextt;
 }e[maxn*];
 void add(int u, int v) {
     e[cnt].to = v;
     e[cnt].nextt = head[u];
     head[u] = cnt++;
 }
 int findd(int u) {
     return f[u] == u?u:f[u] = findd(f[u]);
 }
 void unionn(int x, int y) {
     x = findd(x);
     y = findd(y);
     if(x == y)
         return ;
     f[y] = x;
     int now = (+num[x])/+(num[y]+)/+;
     num[x] = max(now, max(num[x], num[y]));
 }
 int maxx = , pos = -;
 int dfs(int u, int fa, int d) {
     if(d>maxx) {
         maxx = d;
         pos = u;
     }
     for(int i = head[u]; ~i; i = e[i].nextt) {
         int v = e[i].to;
         if(v == fa)
             continue;
         dfs(v, u, d+);
     }
 }
 template<typename __ll>
 inline void read(__ll &m)
 {
     __ll x=,f=;char ch=getchar();
     while(!(ch>=''&&ch<='')){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     m=x*f;
 }
 int main()
 {
     int n, m, q, x, y;
     cin>>n>>m>>q;
     mem1(head);
     for(int i = ; i<=n; i++) {
         f[i] = i;
         num[i] = ;
     }
     while(m--) {
         read(x); read(y);
         add(x, y);
         add(y, x);
         x = findd(x); y = findd(y);
         f[x] = y;
     }
     for(int i = ; i<=n; i++) {
         if(f[i] == i) {
             dfs(i, -, );
             maxx = ;
             if(pos == -)
                 continue;
             dfs(pos, -, );
             num[i] = maxx;
             maxx = ;
             pos = -;
         }
     }
     while(q--) {
         int sign;
         read(sign);
         if(sign == ) {
             read(x);
             printf("%d\n", num[findd(x)]);
         } else {
             read(x); read(y);
             unionn(x, y);
         }
     }
 }