题目：http://acm.hdu.edu.cn/showproblem.php?pid=2586
题意：有n个顶点，n - 1条双向边构成树，每条边都有权值，有m个查询，求每个查询两点之间的最小权值
思路：很久很久以前学最短路时用dijkstra把这题过了。。。用LCA求的话，求出从根节点到每个点之间的权值和，用dist[i]表示，那么对于查询（v, u）就有 dist[v, u] = dist[v] + dist[u] - 2 * dist[LCA(v, u)]，很容易证明这个等式。另外这道题里面的边是双向边，但我建的是单向边，并没有什么影响
总结：居然卡在输出哪里，tarjan求lca，结果并不按查询顺序输出。。。淡淡的忧伤
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 40010;
typedef long long ll;
struct edge
{
int to, cost, next;
}G1[N*2];
struct node
{
int to, ind, next;
}G2[N];
int head1[N], head2[N], par[N];
int cnt1, cnt2;
ll dist[N], res[N];
bool vis[N];
int n, m;
void init()
{
for(int i = 1; i <= n; i++)
par[i] = i;
memset(head1, -1, sizeof head1);
memset(head2, -1, sizeof head2);
memset(vis, 0, sizeof vis);
cnt1 = cnt2 = 0;
}
void add_edge1(int v, int u, int c)
{
G1[cnt1].to = u;
G1[cnt1].cost = c;
G1[cnt1].next = head1[v];
head1[v] = cnt1++;
}
void add_edge2(int v, int u, int ind)
{
G2[cnt2].to = u;
G2[cnt2].ind = ind;
G2[cnt2].next = head2[v];
head2[v] = cnt2++;
}
int ser(int x)
{
int r = x, i = x, j;
while(r != par[r]) r = par[r];
while(i != r) j = par[i], par[i] = r, i = j;
return r;
}
void tarjan_lca(int v)
{
vis[v] = true;
int u;
for(int i = head1[v]; i != -1; i = G1[i].next)
if(!vis[u = G1[i].to])
{
dist[u] = dist[v] + G1[i].cost;
tarjan_lca(u);
par[u] = v;
}
for(int i = head2[v]; i != -1; i = G2[i].next)
if(vis[u = G2[i].to])
res[G2[i].ind] = dist[v] + dist[u] - 2 * dist[ser(u)];
}
int main()
{
int t, a, b, c;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &n, &m);
init();
for(int i = 0; i < n - 1; i++)
{
scanf("%d%d%d", &a, &b, &c);
add_edge1(a, b, c);
add_edge1(b, a, c);
}
for(int i = 0; i < m; i++)
{
scanf("%d%d", &a, &b);
add_edge2(a, b, i);
add_edge2(b, a, i);
}
dist[1] = 0;
tarjan_lca(1);
for(int i = 0; i < m; i++)
printf("%lld\n", res[i]);
}
return 0;
}
