Light-oj-1094 Farthest Nodes in a Tree (树的直径模板题)

Farthest Nodes in a Tree

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Description

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

2

4

0 1 20

1 2 30

2 3 50

5

0 2 20

2 1 10

0 3 29

0 4 50

Sample Output

Case 1: 100

Case 2: 80

AC代码：

#include <stdio.h>
#include<string.h>
#include<queue>
#include <algorithm>
#define MAX 30010
using namespace std;
struct Edge
{
int from, to, val, next;
} edge[MAX*2]; //边的结构体，有正反两个方向，所以要*2；
int head[MAX];
int edgenum; //边的数目；
int dist[MAX]; //统计目前的最长路径；
bool vis[MAX]; //标记统计过的点；
int node, ans;
int N;
int k = 1;
void init()
{
edgenum = 0;
memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int w)
{
edge[edgenum].from = u;
edge[edgenum].to = v;
edge[edgenum].val = w;
edge[edgenum].next = head[u];
head[u] = edgenum++;
}
void BFS(int s)
{
memset(dist, 0, sizeof(dist));
memset(vis, false, sizeof(vis));
queue<int> Q;
Q.push(s);
vis[s] = true;
dist[s] = 0;
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])
{
if(dist[v] < dist[u] + edge[i].val)
{
dist[v] = dist[u] + edge[i].val;
}
vis[v] = true;
Q.push(v);
}
}
}
for(int i = 1; i<=N; i++) {
if(ans < dist[i])
{
ans = dist[i];
node = i;
}
}
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
init();
scanf("%d", &N);
int a, b, c;
for(int i = 1; i < N; i++)
{
scanf("%d%d%d", &a, &b, &c);
a++, b++;
addEdge(a, b, c);
addEdge(b, a, c);
}
BFS(1);
BFS(node);
printf("Case %d: ", k++);
printf("%d\n", ans);
}
return 0;
}