Hdoj 2586 How far away ？ 【LCA】

How far away ？

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

Total Submission(s): 7072 Accepted Submission(s): 2575

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

#include <cstdio>
#include <iostream>
#include <vector>
const int M = 4e4+5;
using namespace std;

struct node{
    int to, val;
};

vector<node > map[M];
vector<int> q[M], num[M];
bool vis[M];
int dis[M], ans[300], fat[M];

int f(int x){
    if(x != fat[x]) fat[x] = f(fat[x]);
    return fat[x];
}

void Lca(int u){
    vis[u] = 1;
    for(int i = 0; i < map[u].size(); ++i){
        int v = map[u][i].to;
        if(!vis[v]){
            dis[v] = dis[u]+map[u][i].val;
            Lca(v);
            fat[v] = u;
        }
    }
    for(int i = 0; i < q[u].size(); ++ i){
        int v = q[u][i];
        if(vis[v]){
            ans[num[u][i]] = dis[u]+dis[v]-2*dis[f(v)];
        }
    }
}

int main(){
    int t, n, m;
    //ios::sync_with_stdio(false);
    scanf("%d", &t);
    while(t --){
      scanf("%d%d", &n, &m);
      for(int i = 1; i <= n; ++ i){
        map[i].clear(); q[i].clear();
        fat[i] = i; vis[i] = 0;
      }
      int a, b, c;
      for(int i = 0; i < n-1; ++i){
        scanf("%d%d%d", &a, &b, &c);
        node temp;
        temp.to = a; temp.val = c;
        map[b].push_back(temp);
        temp.to = b;
        map[a].push_back(temp);
      }
      for(int i = 0; i < m; ++i){
        scanf("%d%d", &a, &b);
        q[a].push_back(b);
        q[b].push_back(a);
        num[a].push_back(i);
        num[b].push_back(i);
      }
      //memset(vis, 0, sizeof(vis));
      dis[1] = 0;
      Lca(1);
      for(int i = 0; i < m; ++ i) printf("%d\n", ans[i]);
    }
    return 0;
}

附RE代码：

#include <cstdio>
#include <iostream>
#include <vector>
const int M = 4e4+5;
using namespace std;

struct node{
    int to, val;
};

vector<node > map[M];
vector<int> q[M], num[M];
bool vis[M];
int dis[M], ans[300], fat[M];

int f(int x){
    if(x != fat[x]) fat[x] = f(fat[x]);
    return fat[x];
}

void Lca(int u, int cur){ //就是这里多了一个参数，溢出了。。
    vis[u] = 1;
    dis[u] = cur;
    for(int i = 0; i < map[u].size(); ++i){
        int v = map[u][i].to;
        if(!vis[v]){
            Lca(v, cur+map[u][i].val);
            fat[v] = u;
        }
    }
    for(int i = 0; i < q[u].size(); ++ i){
        int v = q[u][i];
        if(vis[v]){
            ans[num[u][i]] = dis[u]+dis[v]-2*dis[f(v)];
        }
    }
}

int main(){
    int t, n, m;
    //ios::sync_with_stdio(false);
    scanf("%d", &t);
    while(t --){
      scanf("%d%d", &n, &m);
      for(int i = 1; i <= n; ++ i){
        map[i].clear(); q[i].clear();
        fat[i] = i; vis[i] = 0;
      }
      int a, b, c;
      for(int i = 0; i < n-1; ++i){
        scanf("%d%d%d", &a, &b, &c);
        node temp;
        temp.to = a; temp.val = c;
        map[b].push_back(temp);
        temp.to = b;
        map[a].push_back(temp);
      }
      for(int i = 0; i < m; ++i){
        scanf("%d%d", &a, &b);
        q[a].push_back(b);
        q[b].push_back(a);
        num[a].push_back(i);
        num[b].push_back(i);
      }
      //memset(vis, 0, sizeof(vis));
      Lca(1, 0);
      for(int i = 0; i < m; ++ i) printf("%d\n", ans[i]);
    }
    return 0;
}