UVA 247 Calling Circles (Floyd)

题意:如果两个人互相打电话(直接或者间接),则说他们在同一个电话圈里。例如,a打给b,b打给c,c打给d,d打给a,则这四个人在同一个圈里;如果e打给f,而f不打给e,则不能推出e和f在同一个电话圈。输入n(n<=25)个人的m次电话,找出所有的电话圈。人名只包含字母,不超过25个字符,且不重复。

分析:先由Floyd求出传递闭包,当G[i][j] = G[j][i] = 1 , 说明两者处于同一电话圈。构造一个新图,在一个电话圈里的两个人之间连一条边,然后依次输出各个连通分量的所有人即可

代码:

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <iomanip>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <sstream>
#define ll long long
#define mod 1000000007
#define mem(a) memset(a,0,sizeof(a))

using namespace std;

const int maxn = 30 + 5 , inf = 0x3f3f3f3f ;
int G[maxn][maxn],newG[maxn][maxn],vis[maxn];
map<string,int>a;
map<int,string>b;
void Init(){
    a.clear();
    mem(G);mem(newG);mem(vis);
}
int n,m,k=1,cnt;
void dfs(int u){
    vis[u] = 1;
    if(cnt++) printf(", ");
    cout<<b[u];
    for(int i = 1 ; i <= n ; i ++ ){
        if(!vis[i]&&newG[u][i]&&G[i][u])
            dfs(i);
    }
}
main(){
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    string s1,s2;
    while(~scanf("%d %d",&n,&m),(n||m)){
        Init();
        int Cnt = 1;
        for(int i = 1 ; i <= m ;i ++ ){
            cin>>s1>>s2;
            if(a.find(s1)==a.end()){
                a[s1] = Cnt;
                b[Cnt++] = s1;
            }
            if(a.find(s2)==a.end()){
                a[s2] = Cnt;
                b[Cnt++] = s2;
            }
            G[a[s1]][a[s2]] = 1;
        }
        for(int k = 1 ; k <= n ; k ++ )
            for(int i = 0 ; i <= n ; i ++ )
                for(int j = 0 ; j <= n ; j ++ )
                    if(G[i][k]&&G[k][j]) G[i][j] = 1 ;
        for(int i = 1 ; i <= n ; i ++ )
        for(int j = 1 ; j <= n ; j ++ ){
            if(G[i][j]&&G[j][i]) newG[i][j] = 1 ;
        }
        if(k>1) printf("\n");
        printf("Calling circles for data set %d:\n",k++);
        for(int i = 1 ; i <= n ; i ++ ){
            if(!vis[i]){
                cnt = 0;
                dfs(i);
                printf("\n");
            }
        }
    }
    return 0;
}

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转载自blog.csdn.net/insist_77/article/details/80657473