1021. Deepest Root (25)

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N-1 lines follow, each describes an edge by given the two adjacent nodes’ numbers.

Output Specification:

For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print “Error: K components” where K is the number of connected components in the graph.

Sample Input 1:
5
1 2
1 3
1 4
2 5
Sample Output 1:
3
4
5
Sample Input 2:
5
1 3
1 4
2 5
3 4
Sample Output 2:
Error: 2 components

算法分析：采用领接矩阵存储图，用两次深度优先搜索遍历图，第一次遍历是判断给定的图是否是连通图，如果不是则计算连通图的个数然后输出‘错误’，随便找出最深的顶点；第二次遍历是以第一次遍历中找到的顶点其中任意一个为出发点，再找出最深的顶点。结合前两次遍历中找到的最深的顶点，按从小到大输出。

#include <stdio.h>#include <stdlib.h>int g[10001][10001] = {{0}}; //图//depth表示每个顶点的深度，第一个顶点深度为1；result表示符合要求顶点的集合。int visited[10001] = {0}, depth[10001] = {0}, result[10001] = {0};int n;void dfs(int v, int level) //深度优先遍历，并更新每个顶点的深度{    int i;    visited[v] = 1;    depth[v] = level;    for(i = 1; i <= n; i++)    {        if(!visited[i] && g[v][i] == 1)        {            dfs(i, level + 1);        }    }}void mark() //找出最深的顶点，并初始化{    int maxDepth = -1, i;    for(i = 1; i <= n; i++) //找到最深顶点的深度        if(depth[i] > maxDepth)            maxDepth = depth[i];    for(i = 1; i <= n; i++) //找到最深的顶点，并记录该顶点    {        if(depth[i] == maxDepth)            result[i] = 1;    }    for(i = 1; i <= n; i++) //初始化    {        depth[i] = 0;        visited[i] = 0;    }}int main(){    int i, count = 0;    scanf("%d", &n);    for(i = 0; i < n - 1; i++)    {        int v, w;        scanf("%d %d", &v, &w);        g[v][w] = g[w][v] = 1;    }    for(i = 1; i <= n; i++) //第一次遍历    {        if(!visited[i])        {            dfs(i, 1);            count++;        }    }    if(count > 1) //不是连通图        printf("Error: %d components", count);    else //是连通图    {        mark();         for(i = 1; i <= n; i++) 第二次遍历        {            if(result[i])            {                dfs(i, 1);                break;            }        }        mark();        for(i = 1; i <= n; i++)            if(result[i])                printf("%d\n", i);    }    return 0;}