BFS 3517. The longest athletic track

解题思路：

先随便确定一个顶点，则从该顶点广搜到的最远的点一定是该图的一个起始点，则再从该点出发，广搜。

After a long time of algorithm training, we want to hold a running contest in our beautiful campus. Because all of us are curious about a coders's fierce athletic contest,so we would like a more longer athletic track so that our contest can last more .

In this problem, you can think our campus consists of some vertexes connected by roads which are undirected and make no circles, all pairs of the vertexes in our campus are connected by roads directly or indirectly, so it seems like a tree, ha ha.

We need you write a program to find out the longest athletic track in our campus. our athletic track may consist of several roads but it can't use one road more than once.

Input

*Line 1: A single integer: T represent the case number T <= 10
For each case
*Line1: N the number of vertexes in our campus 10 <= N <= 2000
*Line2~N three integers a, b, c represent there is a road between vertex a and vertex b with c meters long
1<= a,b <= N,  1<= c <= 1000;

Output

For each case only one integer represent the longest athletic track's length

Sample Input

171 2 202 3 102 4 204 5 105 6 104 7 40

Sample Output

80

#include<iostream>#include<cstring>using namespace std;const int num=2000;int n,visit[num],map[num][num],x,y,maxx;struct{int a,sum;}q[num];int bfs(int begin){int max=0,top=1,k;memset(visit,0,sizeof(visit));visit[begin]=1;q[0].a=begin;q[0].sum=0;maxx=0;for(int i=0;i<top;i++){int temp=q[i].a;//i为当前点在栈中所处位置for(int j=1;j<=n;j++)//j为从当前点出发要入栈的点的num{if(map[temp][j]!=-1&&!visit[j]){visit[j]=1;q[top].a=j;//top++;q[top].sum=q[i].sum+map[temp][j];if(q[top].sum>maxx){maxx=q[top].sum;k=j;}top++;}}}return k;}int main(){int t,a,b,c;cin>>t;while(t--){cin>>n;memset(map,-1,sizeof(map));for(int i=0;i<n-1;i++){cin>>a>>b>>c;map[a][b]=map[b][a]=c;}int v=bfs(1);v=bfs(v);cout<<maxx<<endl;}return 0;}