Luogu 4234 最小差值生成树 - LCT 维护链信息

Solution

将边从小到大排序， 添新边$(u, v)$时 若$u,v$不连通则直接添， 若连通则 把链上最小的边去掉 再添边。

若已经加入了 $N - 1$条边则更新答案。

Code

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 #define rd read()

 using namespace std;

 const int N = 1e5 + ;

 const int inf = 1e9;

 int n, m, ans = inf;

 int vis[N << ];

 struct edge {

     int u, v, w;

 }e[N << ];

 int cmp(const edge &A, const edge &B) {

     return A.w < B.w;

 }

 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 cmin(int &A, int B) {

     if (A > B)

         A = B;

 }

 namespace LCT {

     int val[N << ], ch[N << ][], f[N << ], mx[N << ], tun[N << ];

 #define lc(x) ch[x][0]

 #define rc(x) ch[x][1]

     int isroot(int x) {

         return lc(f[x]) != x && rc(f[x]) != x;

     }

     int get(int x) {

         return rc(f[x]) == x;

     }

     void up(int x) {

         mx[x] = val[x];

         if (e[mx[lc(x)]].w < e[mx[x]].w) mx[x] = mx[lc(x)];

         if (e[mx[rc(x)]].w < e[mx[x]].w) mx[x] = mx[rc(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] = ;

         }

     }

 int st[N << ], tp;

     void pd(int x) {

         while (!isroot(x)) {

             st[++tp] = x;

             x = f[x];

         }

         pushdown(x);

         while (tp) pushdown(st[tp--]);

     }

     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[old] = x; f[x] = oldf; 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 main()

 {

     n = rd; m = rd;

     e[].w = inf;

     for (int i = ;  i <= m; ++i) {

         e[i].u = rd; e[i].v = rd; e[i].w = rd;

     }

     sort(e + , e +  + m, cmp);

     for (int i = ; i <= m; ++i)

         val[i + n] = i;

     for (int i = , l = , tot = ; i <= m; ++i) {

         int x = e[i].u, y = e[i].v;

         if (x == y) continue;

         mroot(x);

         if (findr(y) != x) {

             link(i + n, y);

             link(i + n, x);

             vis[i] = ;

             tot++;

             if (tot == n - ) {

                 while (!vis[l]) l++;

                 cmin(ans, e[i].w - e[l].w);

             }

         }

         else {

             int t = mx[y];

             cut(t + n, e[t].u);

             cut(t + n, e[t].v);

             link(i + n, x);

             link(i + n, y);

             vis[t] = ; vis[i] = ;

             if (tot == n - ) {

                 while (!vis[l]) l++;

                 cmin(ans, e[i].w - e[l].w);

             }

         }

     }

     printf("%d\n", ans);

 }