ural1471 Distance in the Tree

最新推荐文章于 2020-02-08 20:01:02 发布

转载最新推荐文章于 2020-02-08 20:01:02 发布 · 131 阅读

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原文链接：http://www.cnblogs.com/dyzll/p/5787081.html

本文介绍了一种求解加权树中两个节点之间距离的算法，并提供了完整的C++实现代码。采用Tarjan算法进行计算，适用于处理大量查询的情况。

Distance in the Tree

Time limit: 1.0 second
Memory limit: 64 MB

A weighted tree is given. You must find the distance between two given nodes.

Input

The first line contains the number of nodes of the tree n (1 ≤ n ≤ 50000). The nodes are numbered from 0 to n – 1. Each of the next n – 1 lines contains three integers u, v, w, which correspond to an edge with weight w (0 ≤ w ≤ 1000) connecting nodes u and v. The next line contains the number of queries m (1 ≤ m ≤ 75000). In each of the next m lines there are two integers.

Output

For each query, output the distance between the nodes with the given numbers.

Sample

input	output
3 1 0 1 2 0 1 3 0 1 0 2 1 2	1 1 2

input

output

1
1
2

分析：树上两点间距离，直接tarjan；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,k,t,p[maxn],ans[maxn],vis[maxn],q[maxn];
int find(int x)
{
    return p[x]==x?x:p[x]=find(p[x]);
}
vi query[maxn],a[maxn],len[maxn],id[maxn];
void dfs(int now,int pre,int gao)
{
    if(pre!=-1)q[now]=q[pre]+gao;
    for(int i=0;i<a[now].size();i++)
    {
        if(a[now][i]!=pre)
        {
            dfs(a[now][i],now,len[now][i]);
        }
    }
}
void dfs1(int now)
{
    vis[now]=1;
    for(int i=0;i<query[now].size();i++)
    {
        if(vis[query[now][i]])
        {
            int fa=find(query[now][i]);
            ans[id[now][i]]=q[now]+q[query[now][i]]-2*q[fa];
        }
    }
    for(int x:a[now])
    {
        if(!vis[x])
        {
            dfs1(x);
            p[x]=now;
        }
    }
}
int main()
{
    int i,j;
    scanf("%d",&n);
    rep(i,1,n)p[i]=i;
    rep(i,1,n-1)
    {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        a[u].pb(v),a[v].pb(u);
        len[u].pb(w),len[v].pb(w);
    }
    scanf("%d",&t);
    rep(i,1,t)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        query[a].pb(b),query[b].pb(a);
        id[a].pb(i),id[b].pb(i);
    }
    dfs(1,-1,-1);
    dfs1(1);
    rep(i,1,t)printf("%d\n",ans[i]);
    //system("Pause");
    return 0;
}