【light oj】树的直径
来源:互联网 发布:0基础学php 编辑:程序博客网 时间:2024/05/16 05:03
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
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384
#include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; struct Edge { int from, to, val, next; }; Edge edge[60000+10]; int head[30000+10];//头'指针' int edgenum;//边数 int dist[30000+10]; //以该点为尾点的最长距离 bool vis[30000+10]; int node; //记录端点值 int ans; //最终的最长路径 int n; int k = 1; void init() //初始化 { edgenum = 0; memset(head, -1, sizeof(head)); } void addedge(int u, int v, int w) { edge[edgenum].from = u;//起点 edge[edgenum].to = v;//终点 edge[edgenum].val = w;//权值 edge[edgenum].next = head[u];//指向下一条边 head[u] = edgenum++; } void BFS(int s) { queue<int> q; memset(dist, 0, sizeof(dist)); memset(vis, false, sizeof(vis)); vis[s] = true; ans = 0; node = s; q.push(s); while(!q.empty()) { int u = q.front(); q.pop(); for(int i = head[u]; i != -1; i = edge[i].next) //遍历每一条边 { int v = edge[i].to; if(!vis[v] && dist[v] < dist[u] + edge[i].val) { vis[v] = true; dist[v] = dist[u] + edge[i].val; if(dist[v] > ans) { ans = dist[v]; node = v; } q.push(v); } } } } int main() { int t; scanf("%d", &t); while(t--) { scanf("%d", &n); init(); 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); //①任意从一个点u出发bfs,设其能到的最远点为v,0不可以 BFS(node); //②从v出发重新bfs,设其能到达的最远点为s printf("Case %d: %d\n", k++, ans); } return 0; }
- 【light oj】树的直径
- light oj 1257 树的直径
- Light OJ--1094(求树的直径)
- light oj 1094 Farthest Nodes in a Tree(树的直径模板)
- Light OJ 1094 - Farthest Nodes in a Tree【树的直径 两次bfs】
- Light oj 1094 - Farthest Nodes in a Tree【树的直径】
- Light OJ 1049 Farthest Nodes in a Tree【树的直径】
- Light OJ 1094 Farthest Nodes in a Tree (树的直径)
- Light oj 1094 Farthest Nodes in a Tree(树的最大直径)
- Light-oj-1094 Farthest Nodes in a Tree (树的直径模板题)
- Light OJ 1049 Farthest Nodes in a Tree(树的直径)(模板题)
- 【Light-oj】-1094 - Farthest Nodes in a Tree(树的直径)
- light oj 1094 Farthest Nodes in a Tree(树的直径模板)
- Light OJ 1094 Farthest Nodes in a Tree(树的直径模板)
- 【Light OJ】1049 Farthest Nodes in a Tree(树的直径模板题)
- Light OJ:1094 Farthest Nodes in a Tree(树状DP+统计树的最大直径)
- XTU OJ Highway(树的直径)
- SDUT OJ 3045 迷之图论 (树的直径)
- 拒绝家电后市场O2O套路,看“找人上门”如何脱颖而出!
- linux命令之ifconfig
- webstorm中console.log的快捷补全方法
- 通过Socket实现TCP编程
- 对深拷贝与浅拷贝的再次理解
- 【light oj】树的直径
- Java Virtual Machine Note
- Arrays.asList剖析
- IOS算法
- 关于在mac系统下GitHub访问不了问题的解决方案
- Set和Map的内部结构
- 牛客网之网易2016实习研发工程师编程题
- 毫秒 时间格式的转换
- C#密码修改及验证