POJ 3694 (tarjan缩点+LCA+并查集)

好久没写过这么长的代码了，题解东哥讲了那么多，并查集优化还是很厉害的，赶快做做前几天碰到的相似的题。

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

 const int N = 1e5 + , M = 2e5 + ;
 int head[N], ver[M * ], Next[M * ];
 int dfn[N], low[N], n, m, tot, num;
 bool bridge[M * ];

 void add(int x, int y)
 {
     ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
 }
 void tarjan(int x, int in_edge)
 {
     dfn[x] = low[x] = ++num;
     for(int i = head[x]; i; i = Next[i])
     {
         int y = ver[i];
         if(!dfn[y])
         {
             tarjan(y, i); ///节点 和 入边
             low[x] = min(low[x], low[y]);

             if(low[y] > dfn[x])
             {
                 bridge[i] = bridge[i ^ ] = true;
             }
         }
         else if(i != (in_edge ^ )) ///搜索树上的边
         {
             low[x] = min(low[x], dfn[y]);
         }
     }
 }
 int c[N], dcc;
 void dfs(int x)
 {
     c[x] = dcc;
     for(int i = head[x]; i; i = Next[i])
     {
         int y = ver[i];
         if(c[y] || bridge[i]) continue;
         dfs(y);
     }
 }
 const int maxn = N;
 ///加边
 int cnt, h[maxn];
 struct edge
 {
     int to, pre, v;
 } e[maxn << ];
 void add(int from, int to, int v)
 {
     cnt++;
     e[cnt].pre = h[from]; ///5-->3-->1-->0
     e[cnt].to = to;
     e[cnt].v = v;
     h[from] = cnt;
 }
 ///LCA
 int dist[maxn];
 int dep[maxn];
 int anc[maxn][]; ///2分的父亲节点
 void dfs(int u, int fa)
 {
     for(int i = h[u]; i; i = e[i].pre)
     {
         int v = e[i].to;
         if(v == fa) continue;
         dist[v] = dist[u] + e[i].v;
         dep[v] = dep[u] + ;
         anc[v][] = u;
         dfs(v, u);
     }
 }
 void LCA_init(int n)
 {
     for(int j = ; ( << j) < n; j++)
         for(int i = ; i <= n; i++) if(anc[i][j-])
         anc[i][j] = anc[anc[i][j-]][j-];
 }
 int LCA(int u, int v)
 {
     int log;
     if(dep[u] < dep[v]) swap(u, v);
     for(log = ; ( << log) < dep[u]; log++);
     for(int i = log; i >= ; i--)
         if(dep[u] - ( << i) >= dep[v]) u = anc[u][i];
     if(u == v) return u;
     for(int i = log; i >= ; i--)
         if(anc[u][i] && anc[u][i] != anc[v][i])
             u = anc[u][i], v = anc[v][i];
     return anc[u][];
 }
 int fa[N];
 int Find(int x)
 {
     if(x == fa[x]) return x;
     return fa[x] = Find(fa[x]);
 }
 int main()
 {
     int kase = ;
     while(scanf("%d %d", &n, &m) != EOF)
     {
         if(n ==  && m == ) break;
         tot = , dcc = , cnt = ;
         for(int i = ; i <= n; i++) head[i] = , dfn[i] = , low[i] = , c[i] = , h[i] = ;
         for(int i = ; i <= m *  + ; i++) ver[i] = , Next[i] = , bridge[i] = false;
         for(int i = ; i <= m; i++)
         {
             int x, y; scanf("%d %d", &x, &y);
             add(x, y), add(y, x);
         }
         ///求桥
         for(int i = ; i <= n; i++)
         {
             if(!dfn[i]) tarjan(i, );
         }
         ///求边双连通分量
         for(int i = ; i <= n; i++)
         {
             if(!c[i])
             {
                 ++dcc;
                 dfs(i);
             }
         }
         ///缩点
         for(int i = ; i <= tot; i++)
         {
             int x = ver[i ^ ], y = ver[i];
             if(c[x] == c[y]) continue;
             add(c[x], c[y], );
         }
         dfs(, );
         LCA_init(dcc);
         ///并查集初始化
         for(int i = ; i <= dcc; i++) fa[i] = i;
         int Q; scanf("%d", &Q);
         int ans = dcc - ;
         printf("Case %d:\n", ++kase);
         while(Q--)
         {
             int x, y; scanf("%d %d", &x, &y);
             x = c[x], y = c[y];
             int p = LCA(x, y);
             x = Find(x);
             while(dep[x] > dep[p])
             {
                 fa[x] = anc[x][];
                 ans--;
                 x = Find(x);
             }
             y = Find(y);
             while(dep[y] > dep[p])
             {
                 fa[y] = anc[y][];
                 ans--;
                 y = Find(y);
             }
             printf("%d\n", ans);
         }
     }
     return ;
 }