hdu 2586 tarjan 板子

hdu 2586

tarjan

#include <cstdio>
#include <cstring>
int const MAX = 40005;

struct Edge
{
int id, val; //当前边序号，边权
int next; //下一条
}e[2 * MAX];

int n, m, cnt;
//x, y表示询问的起点和终点，z是x和y的LCA
int x[MAX], y[MAX], z[MAX];
//fa存祖先，dist存到根的距离，pre存父亲
int fa[MAX], dist[MAX], pre[MAX];
bool vis[MAX];

void addEdge(int u, int v, int w)
{
e[cnt].id = u;
e[cnt].val = w;
e[cnt].next = pre[v];
pre[v] = cnt++;

e[cnt].id = v;
e[cnt].val = w;
e[cnt].next = pre[u];
pre[u] = cnt++;
}

int Find(int x)
{
return x == fa[x] ? x : fa[x] = Find(fa[x]);
}

void tarjan(int k)
{
vis[k] = true;
fa[k] = k;
for(int i = 1; i <= m; i++)
{
if(x[i] == k && vis[y[i]])
z[i] = Find(y[i]);
if(y[i] == k && vis[x[i]])
z[i] = Find(x[i]);
}
for(int i = pre[k]; i != -1; i = e[i].next)
{
if(!vis[e[i].id])
{
dist[e[i].id] = dist[k] + e[i].val;
tarjan(e[i].id);
fa[e[i].id] = k;
}
}
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int u, v, w;
scanf("%d %d", &n, &m);
cnt = 0;
memset(pre, -1, sizeof(pre));
for(int i = 1; i < n; i++)
{
scanf("%d %d %d", &u, &v, &w);
addEdge(u, v, w);
}
for(int i = 1; i <= n; i++)
x[i] = y[i] = z[i] = 0;
for(int i = 1; i <= m; i++)
{
scanf("%d %d", &u, &v);
x[i] = u;
y[i] = v;
}
memset(vis, false, sizeof(vis));
dist[1] = 0;
tarjan(1);
for(int i = 1; i <= m; i++)
printf("%d\n",dist[x[i]] + dist[y[i]] - 2 * dist[z[i]]);
}
}