BZOJ 2594 水管局长 - LCT 维护链信息

Solution

由于链信息不好直接维护， 所以新建一个节点存储边的权值， 并把这个节点连向 它所连的节点 $u$, $v$

$pushup$中更新维护的 $mx$ 指向路径上权值最大的边的编号。

由于这题是只有删边， 没有添边， 所以可以离线倒序， 把删边变成添边。

Code

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

 const int N = 1e6 + ;

 int n, m, Q, ans[N];

 struct edge {
     int u, v, id, val, d;
 }e[N];

 struct que {
     int typ, u, v, id;
 }q[N];

 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;
 }

 namespace LCT {
     int ch[N][], f[N], mx[N], tun[N], val[N];
     int st[N], tp;
 #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 pushdown(int x) {
         if (tun[x]) {
             if (lc(x)) rev(lc(x));
             if (rc(x)) rev(rc(x));
             tun[x] = ;
         }
     }

     void pd(int x) {
         while (!isroot(x)) {
             st[++tp] = x;
             x = f[x];
         }
         st[++tp] = x;
         while (tp) pushdown(st[tp--]);
     }

     void up(int x) {
         mx[x] = val[x];
         if (e[mx[lc(x)]].val > e[mx[x]].val) mx[x] = mx[lc(x)];
         if (e[mx[rc(x)]].val > e[mx[x]].val) mx[x] = mx[rc(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[x] = oldf; f[old] = x; f[son] = old;
         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[x])
             splay(x), ch[x][] = y, up(x);
     }

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

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

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

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

     void cut(int x, int y) {
         split(x, y);
         f[x] = ch[y][] = ;
     }
 }
 using namespace LCT;

 int fd(int x, int y) {
     int l = , r = m;
     for (; l <= r;) {
         int mid = (l + r) >> ;
         if (e[mid].u < x || (e[mid].u == x && e[mid].v < y))
             l = mid + ;
         else if (e[mid].u == x && e[mid].v == y) return mid;
         else r = mid - ;
     }
     return ;
 }

 int cmp1(const edge &A, const edge &B) {
     return A.val < B.val;
 }

 int cmp2(const edge &A, const edge &B) {
     return A.u == B.u ? A.v < B.v : A.u < B.u;
 }

 int cmp3(const edge &A, const edge &B) {
     return A.id < B.id;
 }

 int main()
 {
     n = rd; m = rd; Q = rd;
     for (int i = ; i <= m; ++i) {
         e[i].u = rd, e[i].v = rd, e[i].val = rd;
         if (e[i].u > e[i].v)
             swap(e[i].u, e[i].v);
     }
     sort(e + , e +  + m, cmp1);
     for (int i = ; i <= m; ++i)
         e[i].id = i, val[i + n] = i;
     sort(e + , e +  + m, cmp2);
     for (int i = ; i <= Q; ++i) {
         q[i].typ = rd;
         q[i].u = rd;
         q[i].v = rd;
         if (q[i].u > q[i].v)
             swap(q[i].u, q[i].v);
         if (q[i].typ == ) {
             int t = fd(q[i].u, q[i].v);
             q[i].id = e[t].id;
             e[t].d = ;
         }
     }
     sort(e + , e +  + m, cmp3);
     for (int i = , tot = ; i <= m; ++i) if (!e[i].d) {
         int x = e[i].u, y = e[i].v;
         mroot(x);
         if (findr(y) != x) {
             link(x, i + n);
             link(y, i + n);
             tot++;
         }
         if (tot == n - ) break;
     }
     for (int i = Q; i; i--) {
         if (q[i].typ == ) {
             split(q[i].u, q[i].v);
             ans[i] = e[mx[q[i].v]].val;
         }
         else {
             int x = q[i].u, y = q[i].v;
             mroot(x);
             if (findr(y) != x) {
                 link(y, q[i].id + n);
                 link(x, q[i].id + n);
                 continue;
             }
             if (e[mx[y]].val > e[q[i].id].val) {
                 int t = mx[y];
                 cut(t + n, e[t].u);
                 cut(t + n, e[t].v);
                 link(q[i].id + n, x);
                 link(q[i].id + n, y);
             }
         }
     }
     for (int i = ; i <= Q; ++i)
         if (q[i].typ == )
             printf("%d\n", ans[i]);
 }