hdu 2586 How far away ? Lca的模板了、、
How far away ?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2522 Accepted Submission(s): 931
Problem Description
There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.
Input
First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
Output
For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
Sample Input
2 3 2 1 2 10 3 1 15 1 2 2 3 2 2 1 2 100 1 2 2 1
Sample Output
10 25 100 100
// 差不多就是Lca的模板了、、 dis[i]代表根节点到i的距离
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <stack>
#include <iostream>
#define Max 40001
using namespace std;
struct Node
{
int to;
int next;
__int64 dis;//本来是 LCA
};
int qhead[Max];
Node qedge[522];
int head[Max];
Node edge[Max<<1];
__int64 dis[Max];
int f[Max];
bool visit[Max];
int Find(int x)
{
if(f[x]!=x) f[x]=Find(f[x]);
return f[x];
}
void LCA(int u)
{
visit[u]=true;
int e,v;
for(e=head[u];e!=-1;e=edge[e].next)
{
v=edge[e].to;
if(!visit[v])
{
dis[v]=dis[u]+edge[e].dis;
LCA(v);
f[v]=u;
}
}
for(e=qhead[u];e!=-1;e=qedge[e].next)
{
if(visit[qedge[e].to])
{
int lca=Find(qedge[e].to); //最近公共祖先
qedge[e].dis=dis[u]+dis[qedge[e].to]-(dis[lca]+dis[lca]);
qedge[e^1].dis= qedge[e].dis;
}
}
}
int main()
{
int n,m;
int T;
scanf("%d",&T);
while(T--)
{
int i;
int a,b,k;
int e1=0,e2=0;
scanf("%d %d",&n,&m);
for(i=1;i<=n;i++)
qhead[i]=head[i]=-1;
for(i=1;i<n;i++)
{
scanf("%d %d %d",&a,&b,&k);
edge[e1].to=b;
edge[e1].dis=k;
edge[e1].next=head[a];
head[a]=e1;
e1++;
edge[e1].to=a;
edge[e1].dis=k;
edge[e1].next=head[b];
head[b]=e1;
e1++;
}
for(i=1;i<=m;i++)
{
scanf("%d %d",&a,&b);
qedge[e2].to=b;
qedge[e2].next=qhead[a];
qhead[a]=e2;
e2++;
qedge[e2].to=a;
qedge[e2].next=qhead[b];
qhead[b]=e2;
e2++;
}
for(i=1;i<=n;i++)
f[i]=i,visit[i]=false;
dis[1]=0;
LCA(1);
for(i=0;i<e2;i+=2)
{
printf("%I64d\n",qedge[i].dis);
}
}
return 0;
}