| 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：给定若干两点间的距离，求两点的距离的最大距离。即？树的直径，据说找到离任意一点最远的点index，然后找离index最远的点就是最大距离？
假定0为根节点找离0最远的点就是，最深的点index，然后找离index最远的点就是树上最远的距离
1 #include2 #include 3 #include 4 5 using namespace std; 6 7 #define N 30008 8 9 struct node10 {11 int v, w, next;12 }e[N*4];13 14 int n, cnt, maxx;15 int Index; // 写成index过不了lightoj=·=||16 int head[N], dist[N];17 18 void addedge(int u, int v, int w)19 {20 e[cnt].v = v;21 e[cnt].w = w;22 e[cnt].next = head[u];23 head[u] = cnt++;24 }25 26 void dfs(int u, int w)27 {28 dist[u] = w;29 if(w > maxx)30 {31 maxx = dist[u];32 Index = u;33 }34 for(int i = head[u]; i != -1; i = e[i].next)35 {36 if(dist[e[i].v] == -1)37 {38 dfs(e[i].v, dist[u]+e[i].w);39 }40 }41 }42 int main()43 {44 int t, u, v, w, k = 1;45 46 scanf("%d", &t);47 48 while(t--)49 {50 cnt = maxx = 0;51 memset(head, -1, sizeof(head));52 53 scanf("%d", &n);54 n--;55 while(n--)56 {57 scanf("%d%d%d", &u, &v, &w);58 addedge(u, v, w);59 addedge(v, u, w);60 }61 memset(dist, -1, sizeof(dist));62 dfs(0, 0);63 memset(dist, -1, sizeof(dist));64 dfs(Index, 0);65 printf("Case %d: %d\n", k++, maxx);66 }67 return 0;68 }
bfs
1 #include2 #include 3 #include 4 #include 5 6 using namespace std; 7 8 #define N 30008 9 10 struct node11 {12 int v, w, next;13 } e[N*2];14 15 int n, cnt, maxx;16 int Index;17 int head[N], dist[N], vis[N];18 19 void addedge(int u, int v, int w)20 {21 e[cnt].v = v;22 e[cnt].w = w;23 e[cnt].next = head[u];24 head[u] = cnt++;25 }26 27 void bfs(int u)28 {29 memset(vis, 0, sizeof(vis));30 queue Q;31 Q.push(u);32 vis[u] = 1;33 dist[u] = 0;34 35 while(Q.size())36 {37 u = Q.front();38 Q.pop();39 40 for(int i = head[u]; i != -1; i = e[i].next)41 {42 int v = e[i].v;43 if(!vis[v])44 {45 vis[v] = 1;46 dist[v] = dist[u]+e[i].w;47 if(dist[v] > maxx)48 {49 maxx = dist[v];50 Index = v;51 52 }53 Q.push(v);54 }55 }56 }57 }58 59 int main()60 {61 int t, u, v, w, k = 1;62 scanf("%d", &t);63 64 while(t--)65 {66 maxx = cnt = 0;67 memset(head, -1,sizeof(head));68 69 scanf("%d", &n);70 n--;71 72 while(n--)73 {74 scanf("%d%d%d", &u, &v, &w);75 addedge(u, v, w);76 addedge(v, u, w);77 }78 memset(dist, 0, sizeof(dist));79 bfs(0);80 bfs(Index);81 printf("Case %d: %d\n", k++, maxx);82 }83 return 0;84 }