HDOJ2586 lca查询 tarjan模板程序

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 17516    Accepted Submission(s): 6786


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.

 

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

 

Source

ECJTU 2009 Spring Contest

 

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lcy   |   We have carefully selected several similar problems for you:  3486 2874 2888 3234 2818 

好像离线的算法效率比较高？？？

然后我就用了离线的tarjan算法了。。。

#include <iostream>
#include <ctime>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;

typedef pair<int,int>P;
const int maxn = 4e4+10;
const int maxm = 205;
int father[maxn],dis[maxn],ans[maxm];
int i,n,m;
bool vis[maxn];
vector<P> e[maxn],q[maxn];

void init(){
int x,y,z;
cin >> n >> m;
for (i=0; i<=n; i++) {
e[i].clear();
q[i].clear();
vis[i] = 0;
father[i] = i;
dis[i] = 0;
}

for (i=1; i<n; i++) {
cin >> x >> y >> z;
e[x].push_back(make_pair(y,z));
e[y].push_back(make_pair(x,z));
}

for (i=0; i<m; i++) {
cin >> x >> y;
if (x==y) continue;
q[x].push_back(make_pair(y,i));
q[y].push_back(make_pair(x,i));
}
}

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

void Union(int x, int y){
int fx = Find(x), fy = Find(y);
if (fx!=fy) father[fx] = fy;
}

void tarjan(int x){
int i,v,w,lca;
vis[x] = 1;
for (i=0; i<e[x].size(); i++) {
v = e[x][i].first;
w = e[x][i].second;
if (!vis[v]) {
dis[v] = dis[x] + w;
tarjan(v);
Union(v,x);
father[Find(x)] = x;
}
}

for (i=0; i<q[x].size(); i++) {
v = q[x][i].first;
w = q[x][i].second;
if (vis[v]) {
lca = father[Find(v)];
ans[w] = dis[x] + dis[v] - 2*dis[lca];
}
}
}

int main(){
std::ios::sync_with_stdio(false);
int T;
cin >> T;
while (T--){
init();
tarjan(1);
for (i=0; i<m; i++) cout << ans[i] << endl;
}
return 0;
}