HDU 2586

http://acm.hdu.edu.cn/showproblem.php?pid=2586

题意：求最近祖先节点的权值和

思路：LCA Tarjan算法

#include <stdio.h>
#include <string.h>
#define maxn 40005

int m,n,x[maxn],y[maxn],z[maxn],head[maxn*2],pos,dist[maxn],f[maxn];   
bool vis[maxn];
struct Edge{
    int to,val,next;
}edge[maxn*2];


void add(int u,int v,int val)
{
    edge[pos].to = v;
    edge[pos].val = val;
    edge[pos].next = head[u];
    head[u] = pos++;
}

int Find(int x)
{
    if(f[x]!=x)
        return f[x] =Find(f[x]);
    return f[x];
}

void Tarjan(int u)
{
    vis[u] = true;
    f[u] = u;
    for(int i = 1;i<=n;i++)
    {
        if(x[i]==u&&vis[y[i]]) z[i] = Find(y[i]);
        if(y[i]==u&&vis[x[i]]) z[i] = Find(x[i]);
    }
    for(int i =head[u];i!=-1;i=edge[i].next)
    {
        int v = edge[i].to;
        if(!vis[v])
        {
            dist[v] =dist[u]+edge[i].val;
            Tarjan(v);
            f[v] = u;
        }
    }
}

int main()
{
    int t,a,b,c;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&n);
        memset(head,-1,sizeof(head));
        memset(vis,false,sizeof(vis));

        pos = 1;
        for(int i = 1;i<m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
            z[i] = 0;
        }
        z[m] = 0;
        for(int i = 1; i<=n;i++)
        {
            scanf("%d%d",&x[i],&y[i]);  //分别存每一次的查询点。
        }
        dist[1] = 0;
        Tarjan(1);
        for(int i = 1;i<=n;i++)
            printf("%d
",dist[x[i]]+dist[y[i]]-2*dist[z[i]]);
    }
    return 0;
}