Sicily 8543. Trees

8543. Trees

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB

Description

A graph consists of a set of vertices and edges between pairs of vertices. Two vertices are connected if there is a path (subset of edges) leading from one vertex to another, and a connected component is a maximal subset of vertices that are all connected to each other. A graph consists of one or more connected components.

A tree is a connected component without cycles, but it can also be characterized in other ways. For example, a tree consisting of n vertices has exactly n-1 edges. Also, there is a unique path connecting any pair of vertices in a tree.

Given a graph, report the number of connected components that are also trees.

Input

The input consists of a number of cases. Each case starts with two non-negative integers n and m, satisfying n ≤ 500 and m ≤ n(n-1)/2. This is followed by m lines, each containing two integers specifying the two distinct vertices connected by an edge. No edge will be specified twice (or given again in a different order). The vertices are labelled 1 to n. The end of input is indicated by a line containing n = m = 0.

Output

For each case, print one of the following lines depending on how many different connected components are trees (T > 1 below):

Case x: A forest of T trees.Case x: There is one tree.Case x: No trees.      

x is the case number (starting from 1).

Sample Input

6 31 22 33 46 51 22 33 44 55 66 61 22 31 34 55 66 40 0

Sample Output

Case 1: A forest of 3 trees.Case 2: There is one tree.Case 3: No trees.

Problem Source

Rocky

// Problem#: 8543// Submission#: 3379576// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <stdio.h>#include <iostream>#include <vector>#include <string>#include <stack>#include <iomanip>#include <algorithm>#include <queue>#include <functional>#include <map>#include <string.h>#include <math.h>using namespace std;const int MAX_N = 505;int N, M;bool vis[MAX_N];vector<int> E[MAX_N];vector<int> P;int e, p;void dfs(int now) {    int s = E[now].size();    P.push_back(now);    for (int i = 0; i < s; i++) {        if (!vis[E[now][i]]) {            p++;            vis[E[now][i]] = true;            dfs(E[now][i]);        }    }}void calE() {    bool visFrom[MAX_N];    int s = P.size();    for (int i = 0; i < s; i++) visFrom[P[i]] = false;    for (int i = 0; i < s; i++) {        visFrom[P[i]] = true;        int ss = E[P[i]].size();        for (int j = 0; j < ss; j++) {            if (!visFrom[E[P[i]][j]]) e++;        }    }}int main() {    std::ios::sync_with_stdio(false);    int counter = 1;    while (1) {        cin >> N >> M;        if (!N && !M) break;        for (int i = 1; i <= N; i++) {            vis[i] = false;            E[i].clear();        }        for (int i = 0; i < M; i++) {            int f, t;            cin >> f >> t;            E[f].push_back(t);            E[t].push_back(f);        }        int ans = 0;        for (int i = 1; i <= N; i++) {            if (!vis[i]) {                P.clear();                e = 0;                p = 1;                vis[i] = true;                dfs(i);                calE();                if (p == e + 1) ans++;            }        }        if (ans == 0) cout << "Case " << counter++ << ": No trees." << endl;        if (ans == 1) cout << "Case " << counter++ << ": There is one tree." << endl;        if (ans > 1) cout << "Case " << counter++ << ": A forest of " << ans << " trees." << endl;    }    return 0;}