1094 - Farthest Nodes in a Tree
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| 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.
题意： 给出一棵树，n个顶点，n-1 条边，给出每条边的权值，问树上的任意两点的最大距离是多少
题解：
这个问题被称为求解树的直径（这直径不是很形象啊），很久以前见到大神A这道题的时候感觉好厉害，今天自己尝试着写一下，发现原来并不是太复杂，主要是理清思路...
首先，最容易想到的是用求解最短路的思路来求解最长路，思考后发现，题目给出的是一棵树，那么进行遍历的时候，不必要考虑有环的情况，也就不需要更新操作了 其次，明确不论从哪个点进行搜索，距离这个点最远的某个点，肯定是这棵树的最长路的一个端点（不明白的可以划一下图看看） 这就保证了从任意一个点搜到的最远的点绝对是起点，那么可以先确定起点，随后从这个起点再次遍历整棵树，便能更新出整棵树上的最大距离了
/*
http://blog.csdn.net/liuke19950717
*/
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=30005;
int edgenum,head[maxn],sx,ans;
int dist[maxn],vis[maxn];
struct node
{
int to,val,next;
}edge[maxn*10];
void init()
{
edgenum=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
node tp={v,w,head[u]};
edge[edgenum]=tp;
head[u]=edgenum++;
}
void bfs(int st)
{
memset(dist,0,sizeof(dist));
memset(vis,0,sizeof(vis));
queue<int> q;
q.push(st);
dist[st]=0;vis[st]=1;//起点
ans=0;
while(!q.empty())
{
int u=q.front();q.pop();
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(!vis[v])
{
vis[v]=1;
dist[v]=dist[u]+edge[i].val;//不需要更新，直接赋值
if(ans<dist[v])
{
ans=dist[v];//这个是最长距离
sx=v;//距离当前点最远的点的编号
}
q.push(v);
}
}
}
}
int main()
{
int t;
scanf("%d",&t);
for(int k=1;k<=t;++k)
{
int n;init();
scanf("%d",&n);
for(int i=0;i<n-1;++i)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);add(b,a,c);
}
bfs(0);//搜出sx的值，也就是起点的值
bfs(sx);//搜出最远的距离
printf("Case %d: %d\n",k,ans);
}
return 0;
}