本题的考点是LCA最近公共祖先
How far away ？（HDU2586）
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 19934 Accepted Submission(s): 7806
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
我用的是LCA问题的离线算法的Tarjan算法求解的
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N =
40000+
10;
const int M =
220;
int head[N],head1[N],dis[N],LCA[N],father[N];
bool vis[N];
int n,m,cnt;
struct Edge{
int from,to;
int next;
int val;
};
struct Edge1{
int u,v;
int num;
int next;
};
Edge edge[N*
2];
Edge1 edge1[M*
2];
void add_edge(
int u,
int v,
int val){
edge[cnt].from=u;
edge[cnt].to=v;
edge[cnt].val=val;
edge[cnt].next=head[u];
head[u]=cnt++;
}
void add_ans(
int u,
int v,
int num){
edge1[cnt].u=u;
edge1[cnt].v=v;
edge1[cnt].num=num;
edge1[cnt].next=head1[u];
head1[u]=cnt++;
}
int find(
int x){
int r=x;
while(r!=father[r]){
r=father[r];
}
return r;
}
void tarjan(
int k){
vis[k]=
1;
father[k]=k;
for(
int i=head1[k];i!=-
1;i=edge1[i].next){
int v=edge1[i].v;
if(vis[v]){
LCA[edge1[i].num]=find(v);
}
}
for(
int i=head[k];i!=-
1;i=edge[i].next){
int to=edge[i].to;
if(!vis[to]){
dis[to]=dis[k]+edge[i].val;
tarjan(to);
father[to]=k;
}
}
}
int main(){
int T;
scanf(
"%d",&T);
while(T--){
scanf(
"%d%d",&n,&m);
int a,b,c;
cnt=
0;
memset(head,-
1,
sizeof(head));
memset(dis,
0,
sizeof(dis));
for(
int i=
0;i<n-
1;i++){
scanf(
"%d%d%d",&a,&b,&c);
add_edge(a,b,c);
add_edge(b,a,c);
}
dis[
1]=
0;
cnt=
0;
memset(head1,-
1,
sizeof(head1));
for(
int i=
1;i<=m;i++){
scanf(
"%d%d",&a,&b);
add_ans(a,b,i);
add_ans(b,a,i);
}
memset(vis,
0,
sizeof(vis));
tarjan(
1);
for(
int i=
1;i<=m*
2;i+=
2){
a=edge1[i].u;
b=edge1[i].v;
c=edge1[i].num;
printf(
"%d\n",dis[a]+dis[b]-
2*dis[LCA[c]]);
}
}
return 0;
}
