紫书 例题11-4 UVa247 （Floyd判断联通）

Floyd联通， 然后为了输出联通分量而新建一个图， 让互相可以打电话的建立一条边， 然后dfs输出联通分量就ok了。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<map>
#include<vector>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;

const int MAXN = 30;
int d[MAXN][MAXN], g[MAXN][MAXN], vis[MAXN], n, m;
map<string, int> id;
vector<string> word;

int get_id(string s)
{
	if(id.count(s)) return id[s];
	word.push_back(s);
	return id[s] = word.size() - 1; //注意这里要减去1
}

void print(int u, int p)
{
	if(p != 1) printf(", ");
	cout << word[u];
	vis[u] = 1;
	REP(v, 0, n)
		if(g[u][v] && !vis[v])
			print(v, 0);
}

int main()
{
	int kase = 0;
	while(~scanf("%d%d", &n, &m) && n && m)
	{
		if(kase) puts("");
		word.clear(); id.clear();
		memset(d, 0, sizeof(d));
		memset(vis, 0, sizeof(vis));
		memset(g, 0, sizeof(g));

		while(m--)
		{
			string a, b;
			cin >> a >> b;
			d[get_id(a)][get_id(b)] = 1;
		}

		REP(k, 0, n)
			REP(i, 0, n)
				REP(j, 0, n)
					d[i][j] = d[i][j] || (d[i][k] && d[k][j]);

		REP(i, 0, n)
			REP(j, 0, n)
				g[i][j] = (d[i][j] && d[j][i]);

		printf("Calling circles for data set %d:\n", ++kase);
		REP(i, 0, n)
			if(!vis[i])
			{
				print(i, 1);
				puts("");
			}
	}

	return 0;
}