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http://acm.hdu.edu.cn/showproblem.php?pid=2586
There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can‘t visit a place twice) between every two houses. Yout task is to answer all these curious people.
First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
2
3 2
1 2 10
3 1 15
1 2
2 3
2 2
1 2 100
1 2
2 1
10
25
100
100
tarjan离线求LCA。。
#include <bits/stdc++.h>
using namespace std;
const int N = 40010;
struct Tarjan_Lca {
bool vis[N];
vector<pair<int, int>> A, Q[N];
int tot, par[N], ans[210], head[N], dist[N];
struct edge { int to, w, next; }G[N << 1];
inline void init(int n) {
A.clear();
for(int i = 0; i < n + 2; i++) {
par[i] = i;
vis[i] = false;
head[i] = -1;
dist[i] = 0;
Q[i].clear();
}
}
inline void add_edge(int u, int v, int w) {
G[tot] = { v, w, head[u] }, head[u] = tot++;
G[tot] = { u, w, head[v] }, head[v] = tot++;
}
inline void built(int n, int m) {
int u, v, w;
while(n-- > 1) {
scanf("%d %d %d", &u, &v, &w);
add_edge(u, v, w);
}
for(int i = 0; i < m; i++) {
scanf("%d %d", &u, &v);
A.push_back(pair<int, int>(u, v));
Q[u].push_back(pair<int, int>(v, i));
Q[v].push_back(pair<int, int>(u, i));
}
}
inline int find(int x) {
while(x != par[x]) {
x = par[x] = par[par[x]];
}
return x;
}
inline void tarjan(int u, int fa) {
for(int i = head[u]; ~i; i = G[i].next) {
edge &e = G[i];
if(e.to == fa) continue;
dist[e.to] = dist[u] + e.w;
tarjan(e.to, u);
vis[e.to] = true;
par[e.to] = u;
}
for(auto &r: Q[u]) {
if(vis[r.first]) ans[r.second] = find(r.first);
}
}
inline void solve(int n, int m) {
init(n);
built(n, m);
tarjan(1, 1);
for(int i = 0; i < m; i++) {
printf("%d\n", dist[A[i].first] + dist[A[i].second] - 2 * dist[ans[i]]);
}
}
}go;
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
int t, n, m;
scanf("%d", &t);
while(t--) {
scanf("%d %d", &n, &m);
go.solve(n, m);
}
return 0;
}
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原文地址：http://www.cnblogs.com/GadyPu/p/4992920.html
