【洛谷P4234】最小差值生成树
Description
给定一张n个点，m条边的无向图，求出边权最大值和最小值差值最小的生成树
Solution
LCT+并查集
按照最小生成树的思路，先将边按照边权从小到大排序，然后顺序考虑每一条边
如果当前这条边的两个端点没有连通，那么直接连通
如果两个端点已经连通，我们加上这条边会形成一个环，那么为了让“边权最大值和最小值差值”尽可能小，我们可以将这个环上最短的一条边删掉，换成这条边（显然是对的）
维护最小值可以通过LCT实现，连边、短边也是LCT的基本操作
当目前的边已经是一棵树的时候更新答案即可完成
Code
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int N = 50010; 5 const int M = 200010; 6 inline int read() { 7 int ret = 0, op = 1; 8 char c = getchar(); 9 while (!isdigit(c)) { 10 if (c == '-') op = -1; 11 c = getchar(); 12 } 13 while (isdigit(c)) { 14 ret = (ret << 3) + (ret << 1) + c - '0'; 15 c = getchar(); 16 } 17 return ret * op; 18 } 19 struct Edge { 20 int from, to, dis; 21 bool operator <(const Edge &x) const { 22 return dis < x.dis; 23 } 24 } edge[M]; 25 struct LCT { 26 int fa, val, minn, ch[2], tag; 27 } a[N + M]; 28 int n, m, s[N + M], f[N]; 29 int vis[M]; 30 int find(int x) { 31 if (f[x] != x) return f[x] = find(f[x]); 32 return f[x]; 33 } 34 inline int isnroot(int now) { 35 return a[a[now].fa].ch[0] == now || a[a[now].fa].ch[1] == now; 36 } 37 inline int get(int x, int y) { 38 return a[x].val < a[y].val ? x : y; 39 } 40 inline void update(int now) { 41 int l = a[now].ch[0]; 42 int r = a[now].ch[1]; 43 a[now].minn = get(now, get(a[l].minn, a[r].minn)); 44 } 45 inline void rev(int now) { 46 swap(a[now].ch[0], a[now].ch[1]); 47 a[now].tag ^= 1; 48 } 49 inline void pushdown(int now) { 50 if (a[now].tag) { 51 if (a[now].ch[0]) rev(a[now].ch[0]); 52 if (a[now].ch[1]) rev(a[now].ch[1]); 53 a[now].tag = 0; 54 } 55 } 56 void rotate(int x) { 57 int y = a[x].fa; 58 int z = a[y].fa; 59 int xson = a[y].ch[1] == x; 60 int yson = a[z].ch[1] == y; 61 int B = a[x].ch[xson ^ 1]; 62 if (isnroot(y)) a[z].ch[yson] = x; 63 a[x].ch[xson ^ 1] = y; 64 a[y].ch[xson] = B; 65 if (B) a[B].fa = y; 66 a[y].fa = x; 67 a[x].fa = z; 68 update(y); 69 } 70 void splay(int x) { 71 int y = x, z = 0; 72 s[++z] = y; 73 while (isnroot(y)) y = a[y].fa, s[++z] = y; 74 while (z) pushdown(s[z--]); 75 while (isnroot(x)) { 76 y = a[x].fa; 77 z = a[y].fa; 78 if (isnroot(y)) 79 (a[z].ch[0] == y) ^ (a[y].ch[0] == x) ? rotate(x) : rotate(y); 80 rotate(x); 81 } 82 update(x); 83 } 84 void access(int x) { 85 for (register int y = 0; x; y = x, x = a[x].fa) { 86 splay(x); a[x].ch[1] = y; update(x); 87 } 88 } 89 void makeroot(int x) { 90 access(x); 91 splay(x); 92 rev(x); 93 } 94 void link(int i) { 95 makeroot(edge[i].from); 96 a[edge[i].from].fa = i + N; 97 a[i + N].fa = edge[i].to; 98 } 99 void cut(int i) { 100 access(edge[i - N].from); 101 splay(i); 102 a[a[i].ch[1]].fa = a[a[i].ch[0]].fa = 0; 103 a[i].ch[0] = a[i].ch[1] = 0; 104 } 105 int main() { 106 n = read(); m = read(); 107 for (register int i = 0; i <= n; ++i) f[i] = i, a[i].val = 2147483647; 108 for (register int i = 1; i <= m; ++i) { 109 edge[i].from = read(); edge[i].to = read(); edge[i].dis = read(); 110 } 111 sort(edge + 1, edge + m + 1); 112 int sum = 0, ans = 0, k = 1; 113 for (register int i = 1; i <= m; ++i) { 114 a[i + N].val = edge[i].dis; 115 int x = edge[i].from; 116 int y = edge[i].to; 117 if (find(x) != find(y)) { 118 vis[i] = 1; 119 sum++; 120 link(i); 121 f[f[x]] = f[y]; 122 if (sum == n - 1) ans = edge[i].dis - edge[k].dis; 123 } 124 else { 125 if (x == y) continue ; 126 vis[i] = 1; 127 makeroot(x); 128 access(y); splay(y); 129 vis[a[y].minn - N] = 0; 130 while (vis[k] == 0) ++k; 131 cut(a[y].minn); link(i); 132 if (sum == n - 1) ans = min(ans, edge[i].dis - edge[k].dis); 133 } 134 } 135 printf("%d\n", ans); 136 return 0; 137 }
