Connected undirected weighted graph without self-loops and multiple edges is given. Graph contains n vertices and m edges.
For each edge (u,?v) find the minimal possible weight of the spanning tree that contains the edge (u,?v).
The weight of the spanning tree is the sum of weights of all edges included in spanning tree.
Input
First line contains two integers n and m (1?≤?n?≤?2·105,?n?-?1?≤?m?≤?2·105) — the number of vertices and edges in graph.
Each of the next m lines contains three integers ui,?vi,?wi (1?≤?ui,?vi?≤?n,?ui?≠?vi,?1?≤?wi?≤?109) — the endpoints of the i-th edge and its weight.
Output
Print m lines. i-th line should contain the minimal possible weight of the spanning tree that contains i-th edge.
The edges are numbered from 1 to m in order of their appearing in input.
Examples
Input
Copy
5 71 2 31 3 11 4 52 3 22 5 33 4 24 5 4
Output
98118889题解：建最小支撑树，往里面加边u-v，那么容易想到的是ans=minTree-max{u-....-v这条链上的最大值}+va[u-v]。这里和LCA联系上了，记录从两端点向上跳2^k次得到的最大值，直到跳到u和v的公共祖先。
1 #pragma warning(disable:4996)
2 #include<bitset>
3 #include<string>
4 #include<cstdio>
5 #include<cstring>
6 #include<iostream>
7 #include<algorithm>
8 using namespace std;
9 typedef long long ll;
10
11 const int INF = 2e9 + 9;
12 const int maxn = 200005;
13
14 int n, m, tot, root;
15 int head[maxn], pa[maxn][20], dp[maxn], Fa[maxn], as[maxn][20];
16
17 struct node { int u, v, w; } a[maxn], b[maxn];
18 struct mode { int to, va, next; } e[2*maxn];
19
20 bool cmp(node& x, node& y) { return x.w < y.w; }
21
22 void Inite() {
23 tot = 0;
24 memset(head, -1, sizeof(head));
25
26 memset(as, 0, sizeof(as));
27 memset(dp, 0, sizeof(dp));
28 memset(pa, -1, sizeof(pa));
29 for (int i = 1; i <= n; i++) Fa[i] = i;
30 }
31
32 void addedge(int u, int v, int w) {
33 e[tot].to = v;
34 e[tot].va = w;
35 e[tot].next = head[u];
36 head[u] = tot++;
37 }
38
39 int Find(int a) {
40 if (a == Fa[a]) return a;
41 return Fa[a] = Find(Fa[a]);
42 }
43
44 bool Union(int a, int b) {
45 int x = Find(a), y = Find(b);
46 if (x == y) return false;
47 else {
48 Fa[x] = y;
49 return true;
50 }
51 }
52
53 ll minTree() {
54 sort(a + 1, a + m + 1, cmp);
55 ll ans = 0;
56 for (int i = 1; i <= m; i++) {
57 if (Union(a[i].u, a[i].v)) {
58 ans += a[i].w;
59 root = a[i].u;
60 addedge(a[i].u, a[i].v, a[i].w);
61 addedge(a[i].v, a[i].u, a[i].w);
62 }
63 }
64 return ans;
65 }
66
67 void DFS(int u, int p, int deep) {
68 pa[u][0] = p;
69 dp[u] = deep;
70 for (int i = head[u]; i != -1; i = e[i].next) {
71 int v = e[i].to;
72 if (v == p) continue;
73 as[v][0] = e[i].va;
74 DFS(v, u, deep + 1);
75 }
76 }
77
78 void getPa() {
79 DFS(root, -1, 0);
80 for (int i = 1; (1 << i) < n; i++) {
81 for (int j = 1; j <= n; j++) {
82 if (pa[j][i - 1] != -1) {
83 pa[j][i] = pa[pa[j][i - 1]][i - 1];
84 as[j][i] = max(as[j][i - 1], as[pa[j][i - 1]][i - 1]);
85 }
86 }
87 }
88 }
89
90 int Lca(int u, int v) {
91 if (dp[u] > dp[v]) swap(u, v);
92 int ans = 0;
93 for (int i = 0; i <= 19; i++) if ((dp[v] - dp[u]) >> i & 1) { ans = max(ans, as[v][i]); v = pa[v][i]; }
94 if (u == v) return ans;
95 for (int i = 19; i >= 0; i--) {
96 if (pa[u][i] != pa[v][i]) {
97 int tmp = max(as[u][i], as[v][i]);
98 ans = max(ans, tmp);
99 u = pa[u][i];
100 v = pa[v][i];
101 }
102 }
103 int tmp = max(as[u][0], as[v][0]);
104 ans = max(ans, tmp);
105 return ans;
106 }
107
108 int main()
109 {
110 while (scanf("%d%d", &n, &m) != EOF) {
111 Inite();
112 for (int i = 1; i <= m; i++) scanf("%d%d%d", &a[i].u, &a[i].v, &a[i].w);
113 for (int i = 1; i <= m; i++) b[i].u = a[i].u, b[i].v = a[i].v, b[i].w = a[i].w;
114
115 ll ans1 = minTree();
116 //cout << "root" << " " << root << endl;
117
118 getPa();
119 for (int i = 1; i <= m; i++) {
120 int ans2 = Lca(b[i].u, b[i].v);
121 printf("%I64d\n", ans1 + b[i].w - ans2);
122 }
123 }
124 return 0;
125 }
原文地址：https://www.cnblogs.com/zgglj-com/p/8735347.html