浙大 PAT Advanced level 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

将不含环的图作为一颗树，求让树高度最大的所有根节点。求树的高度很明显可以用bfs。

可以遍历图的每一个节点求高度，然后输出所有高度最大的节点。但需要注意case中有许多稀疏图，只有用邻接表的方式才可以通过所有case，用二维数组表示树在遍历过程中会超时。

实际上更简单的想法是，首先选取任一一个节点做bfs，并记录下使得树高度最高的所有顶点的集合set1，然后从set1中任选一点做bfs，记录下所有使得树高度最高的顶点结合set2，set1和set2的并集即为最终的结果。

/*可以AC，效率很高 0答案正确125613/131答案正确12962/22答案正确23045/53答案正确2315362/24答案正确12562/25答案正确12561/1*/ #include <iostream>#include <vector>#include <queue>#include <set>using namespace std;const int white = 0;const int gray = 1;const int black = 2;typedef struct node{int color;int height;vector<int> adj;}node;int N;vector<node> graph;void dfs(int index){int k;graph[index].color = gray;for (unsigned int i = 0; i < graph[index].adj.size(); ++i){k = graph[index].adj[i];if (white == graph[k].color){dfs(k);}}graph[index].color = black;}void bfs(int root, vector<int> &res){int max_height = 0;int index;int k;queue<int> nq;for (int i = 0; i != N; ++i){graph[i].color = white;graph[i].height = 0;}nq.push(root);graph[root].color = gray;while (!nq.empty()){index = nq.front();nq.pop();graph[index].color = black;for (unsigned int i = 0; i != graph[index].adj.size(); ++i){k = graph[index].adj[i];if (white == graph[k].color){graph[k].height = graph[index].height + 1;graph[k].color = gray;nq.push(k);if (graph[k].height > max_height){max_height = graph[k].height;}}}}for (int i = 0; i != N; ++i){if (graph[i].height == max_height){res.push_back(i);}}return ;}int main(){int ni, nj;vector<int> res1, res2;set<int> res_all;int parts = 0;bool flag = false;cin >> N;graph.resize(N);for (int i = 0; i != N; ++i){graph[i].color = white;graph[i].height = 0;}for (int i = 0; i != N-1; ++i){cin >> ni >> nj;--ni;--nj;graph[ni].adj.push_back(nj);graph[nj].adj.push_back(ni);}for (int i = 0; i != N; ++i){if (white == graph[i].color){dfs(i);++parts;}}if (1 != parts){cout << "Error: " << parts << " components" << endl;return 0;}bfs(0, res1);bfs(res1.at(0), res2);ni = nj = 0;for (unsigned int i = 0; i != res1.size(); ++i){res_all.insert(res1[i]);}for (unsigned int i = 0; i != res2.size(); ++i){res_all.insert(res2[i]);}for (set<int>::const_iterator iter = res_all.begin(); iter != res_all.end(); ++iter){cout << (*iter)+1 << endl;}return 0;}/* 可以ac0答案正确125613/131答案正确12562/22答案正确13005/53答案正确98412122/24答案正确12562/25答案正确13161/1*//* 稀疏图用邻接表效率更高*/#if 0#include <iostream>#include <vector>#include <queue>using namespace std;const int white = 0;const int gray = 1;const int black = 2;typedef struct node{int color;int start;int finish;int height;vector<int> adj;}node;int N;vector<node> graph;int ntime = 0;int parts = 0;void dfs(int index){int k;graph[index].color = gray;graph[index].start = ntime++;for (unsigned int i = 0; i < graph[index].adj.size(); ++i){k = graph[index].adj[i];if (white == graph[k].color){dfs(k);}}graph[index].color = black;graph[index].finish = ntime++;}int bfs(int root){int ret = 0;int index;int k;queue<int> nq;for (int i = 0; i != N; ++i){graph[i].color = white;graph[i].height = 0;}nq.push(root);graph[root].color = gray;while (!nq.empty()){index = nq.front();nq.pop();graph[index].color = black;for (unsigned int i = 0; i != graph[index].adj.size(); ++i){k = graph[index].adj[i];if (white == graph[k].color){graph[k].height = graph[index].height + 1;graph[k].color = gray;nq.push(k);if (graph[k].height > ret){ret = graph[k].height;}}}}return ret;}int main(){int ni, nj, nh;int max_height = 0;vector<int> results;cin >> N;graph.resize(N);for (int i = 0; i != N; ++i){graph[i].color = white;graph[i].height = 0;}for (int i = 0; i != N-1; ++i){cin >> ni >> nj;--ni;--nj;graph[ni].adj.push_back(nj);graph[nj].adj.push_back(ni);}for (int i = 0; i != N; ++i){if (white == graph[i].color){dfs(i);++parts;}}if (1 != parts){cout << "Error: " << parts << " components" << endl;}else{for (int i = 0; i != N; ++i){/************************************************************************//* 注释掉下面的if 语句可以ac*//* 考虑输入:2 1 2*//************************************************************************//* if (graph[i].start + 1 == graph[i].finish) */{nh = bfs(i);if (nh > max_height){max_height = nh;results.clear();results.push_back(i);}else{if (nh == max_height){results.push_back(i);}}}}for (unsigned int i = 0; i != results.size(); ++i){cout << results[i]+1 << endl;}}return 0;}#endif#if 0// 超时#include <iostream>#include <queue>using namespace std;int graph[10002][10002];/* 1表示存在路径*/int visited[10002];/* 1为已经访问*/int result[10002];void DFS(int root, int num){visited[root] = true;for (int i = 1; i <= num; ++i){if (1 == graph[root][i] && 0 == visited[i]){DFS(i, num);}}}// 返回正值表示以root为根的树的最大深度// 返回负值表示图有多少个不连通的部分int levelCount(int root, int num){int level;// 记录以root为根的树的最大深度int part;// 记录图有多少个相互不连通的区域(针对"Error: K components") bool flag = true;// 如果是连通图, 则为true; 反之为falseint count, temp;queue<int> Q;for (int i = 1; i <= num; ++i){visited[i] = 0;}Q.push(root);count = 1;level = 0;visited[root] = 1;part = 1;while(!Q.empty()){while(count--){temp = Q.front();Q.pop();for (int i = 1; i <= num; ++i){if (1 == graph[temp][i] && 0 == visited[i]){visited[i] = 1;Q.push(i);}}}count = Q.size();++level;}for (int i = 1; i <= num; ++i){// 存在未被访问的点if (0 == visited[i]){flag = false;DFS(i, num);++part;}}if (flag){return level;}else{return -part;}}int main(){int num;int row, col;int maxlevel = 0;cin >> num;for (int i = 1; i != num; ++i){cin >> row >> col;graph[row][col] = graph[col][row] = 1;}for (int i = 1; i <= num; ++i){result[i] = levelCount(i, num);if (result[i] < 0){cout << "Error: " << -result[i] <<  " components" << endl;system("pause");return 0;}if (result[i] > maxlevel){maxlevel = result[i];}}for (int i = 1; i <= num; ++i){if (result[i] == maxlevel){cout << i << endl;}}system("pause");return 0;}#endif