题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2874
题目：
Problem Description
After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city. For most of roads had been totally destroyed during the war, there might be no path between two cities, no circle exists as well.
Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.
Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.
Input
Input consists of multiple problem instances.For each instance, first line contains three integers n, m and c, 2<=n<=10000, 0<=m<10000, 1<=c<=1000000. n represents the number of cities numbered from 1 to n. Following m lines, each line has three integers i, j and k, represent a road between city i and city j, with length k. Last c lines, two integers i, j each line, indicates a query of city i and city j.
Output
For each problem instance, one line for each query. If no path between two cities, output “Not connected”, otherwise output the length of the shortest path between them.
Sample Input
5 3 2
1 3 2
2 4 3
5 2 3
1 4
4 5
Sample Output
Not connected
6
思路：用一个vis数组来处理两个节点是否联通，不连通则输出“Not connected”。剩下的联通点之间的距离就是裸的LCA。
代码实现如下：
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std; const int maxn = 1e4 + ;
int n, m, q, u, v, k, cnt;
int cost[maxn], deep[maxn], fa[maxn][], vis[maxn]; struct edge {
int v, l;
edge(int v = , int l = ) : v(v), l(l) {}
}; vector<edge> G[maxn]; void init() {
for(int i = ; i <= n; i++) {
G[i].clear();
}
cnt = ;
memset(vis, , sizeof(vis));
memset(cost, , sizeof(cost));
} void dfs(int u, int d, int p) {
fa[u][] = p;
deep[u] = d;
vis[u] = cnt;
for(int i = ; i < G[u].size(); i++) {
int v = G[u][i].v;
if(v != p) {
cost[v] = cost[u] + G[u][i].l;
dfs(v, d + , u);
}
}
} void lca() {
for(int i = ; i <= n; i++) {
for(int j = ; ( << j) <= n; j++) {
fa[i][j] = -;
}
}
for(int j = ; ( << j) <= n; j++) {
for(int i = ; i <= n; i++) {
if(fa[i][j-] != -) {
fa[i][j] = fa[fa[i][j-]][j-];
}
}
}
} int query(int u, int v) {
if(deep[u] < deep[v]) swap(u, v);
int k;
for(k = ; ( << (k+)) <= deep[u]; k++);
for(int i = k; i >= ; i--) {
if(deep[u] - ( << i) >= deep[v]) {
u = fa[u][i];
}
}
if(u == v) return u;
for(int i = k; i >= ; i--) {
if(fa[u][i] != - && fa[u][i] != fa[v][i]) {
u = fa[u][i];
v = fa[v][i];
}
}
return fa[u][];
} int main() {
while(~scanf("%d%d%d", &n, &m, &q)) {
init();
while(m--) {
scanf("%d%d%d", &u, &v, &k);
G[u].push_back(edge(v, k));
G[v].push_back(edge(u, k));
}
for(int i = ; i <= n; i++) {
if(vis[i] == ) {
cnt++;
dfs(i, , -);
}
}
lca();
for(int i = ; i < q; i++) {
scanf("%d%d", &u, &v);
if(vis[u] != vis[v]) {
printf("Not connected\n");
} else {
printf("%d\n", cost[u] + cost[v] - * cost[query(u,v)]);
}
}
}
return ;
}