BZOJ 1969 航线规划 - LCT 维护边双联通分量

Solution

实际上就是查询 $u$ 到 $v$ 路径上 边双的个数 $ -1$。

并且题目仅有删边， 那么就离线倒序添边。

维护 边双 略有不同：

首先需要一个并查集， 记录 边双内的点。 在 添加边$(u,v)$时 ， 若$u, v$ 已经相连， 那么把 $u, v$ 路径上的点 缩成一个点， 用最上面的点 来代替。

 void del(int x, int y) {
         if (!x) return;
         fa[x] = y;
         del(lc(x), y);
         del(rc(x), y);
     }

压缩点

access 也应该变化 ： $f[x]$ 可能已经经过压缩了。

 void access(int x) {
         for (int y = ; x; y = x, x = f[y] = anc(f[x]))
             splay(x), ch[x][] = y, up(x);
     }

access

Code

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

 const int N = 3e4 + ;
 const int M = 1e5 + ;

 int n, m;
 int fa[N], ans[N], vis[M];

 struct edge {
     int u, v;

     bool operator < (const edge &b) const {
         return u == b.u ? v < b.v : u < b.u;
     }
 }e[M];

 struct que {
     int op, u, v;
 }q[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;
 }

 int anc(int x) {
     return fa[x] == x? x : fa[x] = anc(fa[x]);
 }

 namespace LCT {
     int f[N], sz[N], ch[N][], tun[N];
 #define lc(x) ch[x][0]
 #define rc(x) ch[x][1]

     int isroot(int x) {
         return rc(f[x]) != x && lc(f[x]) != x;
     }

     int get(int x) {
         return rc(f[x]) == x;
     }

     void rev(int x) {
         swap(lc(x), rc(x));
         tun[x] ^= ;
     }

     void up(int x) {
         sz[x] = ;
         if (lc(x)) sz[x] += sz[lc(x)];
         if (rc(x)) sz[x] += sz[rc(x)];
     }

     void pushdown(int x) {
         if (tun[x]) {
             if (lc(x)) rev(lc(x));
             if (rc(x)) rev(rc(x));
             tun[x] = ;
         }
     }

     void pd(int x) {
         if (!isroot(x))
             pd(f[x]);
         pushdown(x);
     }

     void rotate(int x) {
         int old = f[x], oldf = f[old], son = ch[x][get(x) ^ ];
         if (!isroot(old)) ch[oldf][get(old)] = x;
         ch[x][get(x) ^ ] = old;
         ch[old][get(x)] = son;
         f[son] = old; f[x] = oldf; f[old] = x;
         up(old); up(x);
     }

     void splay(int x) {
         pd(x);
         for (; !isroot(x); rotate(x))
             if (!isroot(f[x]))
                 rotate(get(f[x]) == get(x) ? f[x] : x);
     }

     void access(int x) {
         for (int y = ; x; y = x, x = f[y] = anc(f[x]))
             splay(x), ch[x][] = y, up(x);
     }

     int findr(int x) {
         access(x); splay(x);
         while (lc(x)) pushdown(x), x = lc(x);
         return x;
     }

     void mroot(int x) {
         access(x); splay(x); rev(x);
     }

     void split(int x, int y) {
         mroot(x); access(y); splay(y);
     }

     void del(int x, int y) {
         if (!x) return;
         fa[x] = y;
         del(lc(x), y);
         del(rc(x), y);
     }

     void link(int x, int y) {
         mroot(x);
         f[x] = y;
     }

     void merge(int x, int y) {
         x = anc(x); y = anc(y);
         if (x == y)
             return;
         mroot(x);
         if (findr(y) != x)
             link(x, y);
         else {
             splay(x);
             del(rc(x), x);
             rc(x) = ; up(x);
         }
     }
 }

 using namespace LCT;

 int main()
 {
     n = rd; m = rd;
     for (int i = ; i <= n; ++i)
         fa[i] = i;
     for (int i = ; i <= m; ++i) {
         e[i].u = rd; e[i].v = rd;
         if (e[i].u > e[i].v)
             swap(e[i].u, e[i].v);
     }
     sort(e + , e +  + m);
     int Q = ;
     for (; ; ) {
         int op = rd;
         if (op == -)
             break;
         int u = rd, v = rd;
         if (u > v)
             swap(u, v);
         q[++Q].op = op; q[Q].u = u;
         q[Q].v = v;
         edge t;
         t.u = u; t.v = v;
         if (op == )
             vis[lower_bound(e + , e +  + m, t) - e] = ;
     }
     for (int i = ; i <= m; ++i)
         if (!vis[i])
             merge(e[i].u, e[i].v);
     for (int i = Q; i; i--) {
         if (q[i].op == ) {
             int x = q[i].u, y = q[i].v;
             x = anc(x); y = anc(y);
             if (x == y) {ans[i] = ; continue;}
             split(x, y);
             ans[i] = sz[y] - ;
         }
         else merge(q[i].u, q[i].v);
     }
     for (int i = ; i <= Q; ++i)
         if (q[i].op == )
             printf("%d\n", ans[i]);
 }