uva 297(传递闭包 WF 1996)

题意：在一张有向图中输出所有的环。

思路：先用Floyd求传递闭包，然后通过传递闭包建图若是Map[i][j] && Map[j][i]则建一条无向边。然后图中所有的连通分支即为一个环。

代码如下：

/**************************************************
 * Author     : xiaohao Z
 * Blog     : http://www.cnblogs.com/shu-xiaohao/
 * Last modified : 2014-06-26 11:12
 * Filename     : uva_247.cpp
 * Description     : 
 * ************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<unsigned int,unsigned int> puu;
typedef pair<int, double> pid;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

const int INF = 0x3f3f3f3f;
const double eps = 1E-6;
const int LEN = 1010;
vector<int> Map[LEN], ans[LEN];
map<string, int> mp;
string a, b, name[LEN];
int n, m, top, cnt, Mp[LEN][LEN], vis[LEN];

int ch(string a){
    if(!mp.count(a)) {
        name[top] = a;
        mp[a] = top++;
    }
    return mp[a];
}

void Floyd(){
    for(int k=0; k<n; k++){
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(Mp[i][j] || (Mp[i][k] && Mp[k][j])) {
                    Mp[i][j] = 1;
                }
            }
        }
    }
}

void dfs(int v, int tag){
    vis[v] = tag;
    for(int i=0; i<Map[v].size(); i++){
        int x = Map[v][i];
        if(vis[x] < 0) dfs(x, tag);
    }
}

void solve(){
    for(int i=0; i<n; i++){
        for(int j=0; j<n; j++){
            if(Mp[i][j] && Mp[j][i]){
                Map[i].PB(j);
                Map[j].PB(i);
            }
        }
    }
    cnt = 0;
    memset(vis, -1, sizeof vis);
    for(int i=0; i<n; i++){
        if(vis[i] < 0) dfs(i, cnt++);
    }
    for(int i=0; i<cnt; i++){
        for(int j=0; j<n; j++){
            if(vis[j] == i){
                ans[i].PB(j);
            }
        }
    }
}

void out(int &kase){
    if(kase != 1) cout << endl;
    cout << "Calling circles for data set "<< kase++ << ":" << endl;
    for(int i=0; i<cnt; i++){
        for(int j=0; j<ans[i].size(); j++){
            cout << name[ans[i][j]];
            if(j != ans[i].size()-1)cout << ", ";
        }
        cout << endl;
    }
}

int main()
{
//    freopen("in.txt", "r", stdin);

    ios::sync_with_stdio(false);
    int kase = 1;
    while(cin >> n >> m){
        if(!n && !m) break;
        memset(Mp, 0, sizeof Mp);
        for(int i=0; i<LEN; i++) Map[i].clear();
        for(int i=0; i<LEN; i++) ans[i].clear();
        mp.clear();top = 0;
        for(int i=0; i<m; i++){
            cin >> a >> b;
            int na = ch(a), nb = ch(b);
            Mp[na][nb] = 1;
        }
        Floyd();
        solve();
        out(kase);
    }
    return 0;
}