PAT (Advanced Level) Practise 1021 Deepest Root (25)

1021. Deepest Root (25)

时间限制

1500 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

51 21 31 42 5

Sample Output 1:

345

Sample Input 2:

51 31 42 53 4

Sample Output 2:

Error: 2 components

对于不能成树的用并查集来判断，对于能成树的，随便选择一个点，找到最远点，所有最远点都可以是答案，然后随便找一个最远点，以这个点为根再找到全部最远点

那些点也是答案，然后排个序输出即可。至于为什么可以这么做，可以参考树的直径的证明。

#include<cstdio>#include<string>#include<cstring>#include<vector>#include<iostream>#include<queue>#include<algorithm>using namespace std;typedef long long LL;const int INF = 0x7FFFFFFF;const int maxn = 2e4 + 10;vector<int> t[maxn];int n, fa[maxn], x, y, cnt, ans[maxn], dis[maxn], maxd;int get(int x){return x == fa[x] ? x : fa[x] = get(fa[x]);}void dfs(int x, int y, int dep){dis[x] = dep;maxd = max(maxd, dep);for (int i = 0; i < t[x].size(); i++){if (t[x][i] == y) continue;dfs(t[x][i], x, dep + 1);}}int main(){scanf("%d", &n);cnt = n;for (int i = 1; i <= n; i++) fa[i] = i;for (int i = 1; i < n; i++){scanf("%d%d", &x, &y);int fx = get(x), fy = get(y);if (fx != fy) { fa[fx] = fy; cnt--; }t[x].push_back(y);t[y].push_back(x);}if (cnt > 1) printf("Error: %d components\n", cnt);else{cnt = 0;dfs(1, 1, 0);for (int i = 1; i <= n; i++) if (dis[i] == maxd) ans[cnt++] = i;for (int i = 1; i <= n; i++) dis[i] = 0;dfs(ans[0], ans[0], 0);for (int i = 1; i <= n; i++) if (dis[i] == maxd) ans[cnt++] = i;sort(ans, ans + cnt);cnt = unique(ans, ans + cnt) - ans;for (int i = 0; i < cnt; i++) printf("%d\n", ans[i]);}return 0;}