POJ3694 Network - Tarjan + 并查集

Description

给定$N$个点和 $M$条边的无向联通图， 有$Q$ 次操作， 连接两个点的边， 问每次操作后的图中有几个桥

Solution

首先Tarjan找出边双联通分量， 每个双联通分量缩成一个点， 就构成了一棵树， 每一条树边都是桥。

执行连$u, v$ 边时， 用并查集跳到没有桥的深度最浅并且深度比$lca$深的点， 将它与父节点的并查集合并， 再接着跳。

每跳一次， 桥的数量就减少$1$。

另外感谢Iowa 神犇提醒我$cut$数组要开$M << 1$, 不是 $N << 1$, 拯救了$RE$崩溃的我呜呜

Code

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define rd read()
 using namespace std;

 const int N = 1e5 + 1e4;
 const int M = 2e5 + 1e4;

 int n, m, dfn[N], low[N], cnt;
 int head[N], tot;
 int Head[N], Tot;
 int col[N], col_num, father[N], ans;
 int top[N], son[N], size[N], f[N], dep[N];
 int cut[M << ];

 struct edge {
     int nxt, to;
 }E[M << ], e[M << ];

 int read() {
     int X = , p = ; char c = getchar();
     for(; c > '' || c < ''; c = getchar()) if(c == '-') p = -;
     for(; c >= '' && c <= ''; c = getchar()) X = X *  + c - '';
     return X * p;
 }

 void add(int u, int v) {
     e[++tot].to = v;
     e[tot].nxt = head[u];
     head[u] = tot;
 }

 void Add(int u, int v) {
     E[++Tot].to = v;
     E[Tot].nxt = Head[u];
     Head[u] = Tot;
 }

 void dfs1(int u) {
     size[u] = ;
     for(int i = Head[u]; i; i = E[i].nxt) {
         int nt = E[i].to;
         if(nt == f[u]) continue;
         f[nt] = u;
         dep[nt] = dep[u] + ;
         dfs1(nt);
         size[u] += size[nt];
         if(size[nt] > size[son[u]]) son[u] = nt;
     }
 }

 void dfs2(int u) {
     if(!son[u]) return;
     top[son[u]] = top[u];
     dfs2(son[u]);
     for(int i = Head[u]; i; i = E[i].nxt) {
         int nt = E[i].to;
         if(nt == f[u] || nt == son[u]) continue;
         top[nt] = nt;
         dfs2(nt);
     }
 }

 int LCA(int x, int y) {
     for(; top[x] != top[y];) {
         if(dep[top[x]] < dep[top[y]]) swap(x, y);
         x = f[top[x]];
     }
     if(dep[x] < dep[y]) swap(x, y);
     return y;
 }

 int get(int x) {
     return father[x] == x? x : father[x] = get(father[x]);
 }

 void merg(int x, int y) {
     x = get(x); y = get(y);
     father[x] = y;
 }

 int ch(int x) {
     return ((x + ) ^ ) - ;
 }

 void tarjan(int u, int pre) {
     dfn[u] = low[u] = ++cnt;
     for(int i = head[u]; i; i = e[i].nxt) {
         if(i == ch(pre)) continue;
         int nt = e[i].to;
                 if(!dfn[nt]) {
                     tarjan(nt, i);
                     low[u] = min(low[u], low[nt]);
                     if(low[nt] > dfn[u]) {
                         cut[ch(i)] = cut[i] = ;
                         ans++;
                     }
                 }
                 else low[u] = min(low[u], dfn[nt]);
     }
 }

 void dfs(int u) {
     col[u] = col_num;
     for(int i = head[u]; i; i = e[i].nxt) {
         int nt = e[i].to;
         if(col[nt] || cut[i]) continue;
         dfs(nt);
         //blo[col_num].push_back(nt);
     }
 }

 void work() {
     ans = Tot = tot = cnt = col_num = ;
     memset(Head, , sizeof(Head));
     memset(head, , sizeof(head));
     memset(cut, , sizeof(cut));
     memset(dfn, , sizeof(dfn));
     memset(col, , sizeof(col));
     memset(son, , sizeof(son));
     memset(size, , sizeof(size));
     for(int i = ; i <= m; ++i) {
         int u = rd, v = rd;
         add(u, v); add(v, u);
     }
     for(int i = ; i <= n; ++i) if(!dfn[i]) tarjan(i, );
     for(int i = ; i <= n; ++i) if(!col[i]) {
         ++col_num; dfs(i);
     }
     for(int i = ; i <= tot; ++i) {
         int x = e[i].to, y = e[ch(i)].to;
         if(col[x] == col[y]) continue;
         Add(col[x], col[y]);
     }
     for(int i = ; i <= col_num; ++i) father[i] = i;
     dfs1();
     top[] = ; dfs2();
     int T = rd;
     for(; T; T--) {
         int u = col[rd], v = col[rd], lca = LCA(u, v);
         u = get(u); v = get(v);
         while(dep[u] > dep[lca]) {
             merg(u, f[u]);
             u = get(u);
             ans --;
         }
         while(dep[v] > dep[lca]) {
             merg(v, f[v]);
             v = get(v);
             ans --;
         }
         printf("%d\n", ans);
     }
 }

 int main()
 {
     for(int i = ; ; ++i) {
         n = rd; m = rd;
         if(!n && !m) return ;
         printf("Case %d:\n", i);
         work();
         putchar('\n');
     }
 }