| 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-1lines 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.
树的直径裸题
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define MAX 40010
using namespace std;
int head[MAX];
int ans=0,n,m,beg;
int sum;
int dis[MAX],vis[MAX];
struct node
{
int u,v,w;
int next;
}edge[MAX];
void add(int u,int v,int w)
{
edge[ans].u=u;
edge[ans].v=v;
edge[ans].w=w;
edge[ans].next=head[u];
head[u]=ans++;
}
void bfs(int x)
{
queue<int>q;
int i,j;
memset(dis,0,sizeof(dis));
memset(vis,0,sizeof(vis));
while(!q.empty())
q.pop();
beg=x;
vis[beg]=1;
sum=0;
q.push(beg);
int top;
while(!q.empty())
{
top=q.front();
q.pop();
for(i=head[top];i!=-1;i=edge[i].next)
{
int z=edge[i].v;
if(!vis[z])
{
dis[z]=dis[top]+edge[i].w;
vis[z]=1;
q.push(z);
if(sum<dis[z])
{
sum=dis[z];
beg=z;
}
}
}
}
}
int main()
{
int i,j,t;
int a,b,c;
scanf("%d",&t);
while(t--)
{
memset(head,-1,sizeof(head));
scanf("%d",&n);
for(i=1;i<n;i++)
{
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);
add(b,a,c);
}
bfs(0);
bfs(beg);
printf("%d
",sum);
}
return 0;
}