题意：一棵树，多次询问任意两点的路径长度。
LCA：最近公共祖先Least Common Ancestors。两个节点向根爬，第一个碰在一起的结点。
求出x, y的最近公共祖先lca后，假设dist[x]为x到根的距离，那么x->y的距离为dist[x]+dist[y]-2*dist[lca]
求最近公共祖先解法常见的有两种
1, tarjan+并查集
2,树上倍增
首先是树上倍增。
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <algorithm>
5
6 using namespace std;
7 const int maxn = 4e4 + 7;
8
9 int T, n, m, u, v, w;
10
11 int first[maxn], sign, st[maxn][21], level[maxn], dist[maxn];
12
13 struct Node {
14 int to, w, next;
15 } edge[maxn * 2];
16
17 inline void init() {
18 for(int i = 0; i <= n; i ++ ) {
19 first[i] = -1;
20 }
21 sign = 0;
22 }
23
24 inline void add_edge(int u, int v, int w) {
25 edge[sign].to = v;
26 edge[sign].w = w;
27 edge[sign].next = first[u];
28 first[u] = sign ++;
29 }
30
31 void dfs(int now, int father) {
32 level[now] = level[father] + 1;
33 st[now][0] = father;
34 for(int i = 1; (1 << i) <= level[now]; i ++ ) {
35 st[now][i] = st[ st[now][i - 1] ][i - 1];
36 }
37 for(int i = first[now]; ~i; i = edge[i].next) {
38 int to = edge[i].to, w = edge[i].w;
39 if(to == father) {
40 continue;
41 }
42 dist[to] = dist[now] + w;
43 dfs(to, now);
44 }
45 }
46
47 int LCA(int x, int y) {
48 if(level[x] > level[y]) {
49 swap(x, y);
50 }
51 for(int i = 20; i >= 0; i -- ) {
52 if(level[x] + (1 << i) <= level[y]) {
53 y = st[y][i];
54 }
55 }
56 if(x == y) {
57 return x;
58 }
59 for(int i = 20; i >= 0; i -- ) {
60 if(st[x][i] == st[y][i]) {
61 continue;
62 } else {
63 x = st[x][i], y = st[y][i];
64 }
65 }
66 return st[x][0];
67 }
68
69 int main() {
70 scanf("%d", &T);
71 while(T--) {
72 scanf("%d %d", &n, &m);
73 init();
74 for(int i = 1; i <= n - 1; i ++ ) {
75 scanf("%d %d %d", &u, &v, &w);
76 add_edge(u, v, w);
77 add_edge(v, u, w);
78 }
79 memset(level, 0, sizeof(level));
80 memset(st, 0, sizeof(st));
81 dist[1] = 0;
82 dfs(1, 0);
83 for(int i = 1; i <= m; i ++ ) {
84 int u, v;
85 scanf("%d %d", &u, &v);
86 int lca = LCA(u, v);
87 printf("%d\n", dist[u] + dist[v] - 2 * dist[lca]);
88 }
89 }
90
91 return 0;
92 }
然后是tarjan的解法(参考算法竞赛进阶指南)
个人理解，tarjan算法就是对搜索的过程中维护了一些性质。在搜索的过程中把点分为三类
1,已经范围且回溯的点，代码中vis[x]=2
2,已经访问还没有回溯的点,代码中vis[x]=1
3,未被标记的点
每一个回溯的点都用并查集连到他的父节点，这样如果我们x点正在访问，我们需要知道x,y的lca,而y已经被访问了，那么并查集y的根就是x,y的lca
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <algorithm>
5
6 using namespace std;
7 const int MAXN = 5e4 + 7;
8 const int INF = 0x3f3f3f3f;
9
10 int n, m, first[MAXN], sign;
11
12 int pre[MAXN], ans[MAXN], vis[MAXN], dist[MAXN], lca[MAXN], indexs;
13
14 vector<pair<int, int> >query[MAXN]; ///y, id
15
16 struct Node {
17 int to, w, next;
18 } edge[MAXN * 2];
19
20 inline void init() {
21 for(int i = 0; i <= n; i++ ) {
22 first[i] = -1;
23 pre[i] = i;
24 query[i].clear();
25 ans[i] = INF;
26 vis[i] = 0;
27 }
28 sign = 0;
29 }
30
31 inline void add_edge(int u, int v, int w) {
32 edge[sign].to = v;
33 edge[sign].w = w;
34 edge[sign].next = first[u];
35 first[u] = sign++;
36 }
37
38 int findx(int x) {
39 return pre[x] == x ? x : pre[x] = findx(pre[x]);
40 }
41
42 inline void join(int x, int y) {
43 int fx = findx(x), fy = findx(y);
44 pre[fx] = fy;
45 }
46
47 inline bool same(int x, int y) {
48 return findx(x) == findx(y);
49 }
50
51 void tarjan(int now) {
52 vis[now] = 1;
53 for(int i = first[now]; ~i; i = edge[i].next) {
54 int to = edge[i].to;
55 if(!vis[to]) {
56 dist[to] = dist[now] + edge[i].w;
57 tarjan(to);
58 pre[to] = now;
59 }
60 }
61 for(int i = 0; i < query[now].size(); i++ ) {
62 int y = query[now][i].first, id = query[now][i].second;
63 if(vis[y] == 2) {
64 int lca = findx(y);
65 ans[id] = min(ans[id], dist[now] + dist[y] - 2 * dist[lca]);
66 }
67 }
68 vis[now] = 2;
69 }
70
71 int main()
72 {
73 int T;
74 scanf("%d", &T);
75 while(T--) {
76 scanf("%d %d", &n, &m);
77 init();
78 for(int i = 1; i <= n - 1; i++ ) {
79 int u, v, w;
80 scanf("%d %d %d", &u, &v, &w);
81 add_edge(u, v, w);
82 add_edge(v, u, w);
83 }
84 for(int i = 1; i <= m; i++ ) {
85 int x, y;
86 scanf("%d %d", &x, &y);
87 if(x == y) {
88 ans[i] = 0;
89 continue;
90 }
91 query[x].push_back(make_pair(y, i));
92 query[y].push_back(make_pair(x, i));
93 ans[i] = INF;
94 }
95 tarjan(1);
96 for(int i = 1; i <= m; i++ ) {
97 printf("%d\n", ans[i]);
98 }
99 }
100 return 0;
101 }
原文地址：https://www.cnblogs.com/Q1143316492/p/9194214.html
时间： 2024-11-10 18:59:36