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

23 21 2 103 1 151 22 32 21 2 1001 22 1

 

Sample Output

1025100100

 

Source

ECJTU 2009 Spring Contest

 

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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;}