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);
 }