E - Farthest Nodes in a Tree

E - Farthest Nodes in a Tree

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
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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

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#include<cstdio>#include<cstring>#include<queue>using namespace std;const int MAXN=1e6+10;int dis[MAXN],head[MAXN];bool vis[MAXN];int edgenum,Tnode,ans,n;struct Edge{    int from;    int to;    int val;    int next;}edge[MAXN];void init(){    memset(head,-1,sizeof(head));edgenum=0;}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(dis,0,sizeof(dis));    memset(vis,false,sizeof(vis));    queue<int> que;    que.push(s);vis[s]=true;dis[s]=0;//第一个入队的要提前标记     ans=0;    while(!que.empty())    {        int now=que.front();que.pop();//取出节点 清除         for(int i=head[now];i!=-1;i=edge[i].next)        {            int go=edge[i].to;//访问与之相邻接的节点             if(!vis[go])//如果没访问过 访问             {                if(dis[go]<dis[now]+edge[i].val)//更新距离                 dis[go]=dis[now]+edge[i].val;                vis[go]=true;//标记访问过                 que.push(go);//入队             }        }    }    for(int i=0;i<n;++i)    {          if(ans<dis[i])          {             ans=dis[i];             Tnode=i;          }    }}int main(){    int kase=0,t,u,v,w,i;    scanf("%d",&t);    while(t--)    {        init();//初始化         scanf("%d",&n);        for(i=1;i<n;++i)        {            scanf("%d%d%d",&u,&v,&w);            addedge(u,v,w);            addedge(v,u,w);        }        bfs(0);//从某一点开始搜索 搜到最远点         bfs(Tnode)//从最远点在搜索到另一最远点         printf("Case %d: %d\n",++kase,ans);    }    return 0;}