SPOJ_PT07Z 树最长路径问题

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1. 题面

You are given an unweighted, undirected tree. Write a program to output the length of the longest path (from one node to another) in that tree. The length of a path in this case is number of edges we traverse from source to destination.

Input

The first line of the input file contains one integer N — number of nodes in the tree (0 < N <= 10000). Next N-1 lines contain N-1 edges of that tree — Each line contains a pair (u, v) means there is an edge between node u and node v (1 <= u, v <= N).

Output

Print the length of the longest path on one line.

Example

Input:
3
1 2
2 3

Output:
2

2.思路

构造三维数组存储每个节点 ①最长路 ②最长边 ③次长边
使用vector存储相邻结点地址
DFS更新最长路，即为答案

3.代码

#include <iostream>#include <vector>using namespace std;int check[10000 + 5];int treenode[10000 + 5][3];int n, m,ans=1;void dfs(vector<int> *nodelink , int i) {    if (nodelink[i].size() == 1) return;    int n1=0, n2=0;    for (int k = 0; k < nodelink[i].size(); k++) {        check[i]++;        int temp = nodelink[i][k];        if (!check[temp]) {            dfs(nodelink, temp);            if (treenode[i][1] < treenode[temp][1]) {                treenode[i][2] = treenode[i][1]>treenode[i][2]?treenode[i][1]:treenode[i][2];                treenode[i][1] = treenode[temp][1];            }            else {                if (treenode[i][2]<treenode[temp][1])                    treenode[i][2] = treenode[temp][1];            }        }    }    treenode[i][1]++;    treenode[i][2]++;    treenode[i][0] = treenode[i][1] + treenode[i][2];    ans = treenode[i][0] > ans ? treenode[i][0] : ans;}int main() {    while (cin >> n) {        vector<int> nodelink[10000+5];        int u, v;        for (int i = 0; i < n-1; i++) {            cin >> u >> v;            nodelink[u].push_back(v);            nodelink[v].push_back(u);        }        int k = 1;        for (; k < n + 1; k++)            if (nodelink[k].size()>1) break;        dfs(nodelink,k);        cout << ans << endl;        ans = 0;        for (int i = 0; i < n + 1; i++){            treenode[i][0] = 0;            treenode[i][1] = 0;            treenode[i][2] = 0;            check[i] = 0;        }    }    return 0;}

4.所得

题意与PT07Y相似，改进处为用vector存储结点地址

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