lightoj-1094 Farthest Nodes in a Tree(求树的直径)

1094 - Farthest Nodes in a Tree
PDF (English) Statistics Forum
Time Limit: 2 second(s) Memory Limit: 32 MB
Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undirected. That means you have to find two nodes in the tree whose distance is maximum amongst all nodes.

Input
Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with an integer n (2 ≤ n ≤ 30000) denoting the total number of nodes in the tree. The nodes are numbered from 0 to n-1. Each of the next n-1 lines will contain three integers u v w (0 ≤ u, v < n, u ≠ v, 1 ≤ w ≤ 10000) denoting that node u and v are connected by an edge whose weight is w. You can assume that the input will form a valid tree.

Output
For each case, print the case number and the maximum distance.

Sample Input
Output for Sample Input
2
4
0 1 20
1 2 30
2 3 50
5
0 2 20
2 1 10
0 3 29
0 4 50
Case 1: 100
Case 2: 80
Notes
Dataset is huge, use faster i/o methods.

 

思路：从0节点开始dfs出一条最长的路径(因为是求树的直径，肯定是两条最长边，但又不能确定都是经过0的，所以先dfs出其中一条)，以该路径叶节点作为根，再dfs出最长路径。

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

const int N = 30010;
struct edge{
    
    int u,v,w,next;
    edge(){};
    edge(int u,int v,int w,int next):u(u),v(v),w(w),next(next){};
    
}E[2*N];

int head[N],tot,T,n,maxn,pos;

void addEdge(int u,int v,int w){ // 将多叉树转成二叉树 
    E[tot] = edge(u,v,w,head[u]);
    head[u] = tot++;
    E[tot] = edge(v,u,w,head[v]);
    head[v] = tot++;
}

void dfs(int u,int v,int w){ // v 是为了防止往回dfs 会出现无限循环 导致栈溢出 
    int child = 0;
    for(int i=head[u];i!=-1;i = E[i].next){//cout<<i<<endl; 
        if(E[i].v==v) continue;
        dfs(E[i].v,u,E[i].w+w);
        child++;
    }
    
    if(child==0&&w>maxn){
        maxn = w;
        pos = u;//cout<<" "<<w<<endl;
    }
    return ;
}

int main(){
    
    int u,v,w;
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        
        tot = 0;
        memset(head,-1,sizeof(head));
        scanf("%d",&n);
        for(int i=1;i<n;i++){
            scanf("%d%d%d",&u,&v,&w);
            addEdge(u,v,w);
        }
        maxn = 0;
        dfs(0,-1,0);
        maxn = 0;
        //cout<<pos<<endl;
        dfs(pos,-1,0);
        printf("Case %d: %d\n",t,maxn);
    }
    
    
    return 0;
}