UVa 247 Calling Circles (传递闭包)

题目链接:https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=183

求出传递闭包，知道两个人之间能否直接或间接互相联系，如果能互相联系则在一个电话圈里，建新图输出每个连通分量即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 110;

int n, m, tot;
int d[maxn][maxn];
map<string, int> mp;
vector<string> names;

int h[maxn], cnt = 0;
struct Edge{
	int to, next;
}e[maxn * maxn * 2];
void add(int u, int v){
	e[++cnt].to = v;
	e[cnt].next = h[u];
	h[u] = cnt;
}

int ID(const string s){
	if(mp.count(s) != 0){
		return mp[s];
	} else{
		names.push_back(s);
		mp[s] = ++tot;
		return tot;
	}
}

int f = 0;
int vis[maxn];
void dfs(int u){
	if(vis[u]) return;
	if(f) printf(", ");
	f = 1;
	printf("%s", names[u-1].c_str());
	
	vis[u] = 1;
	for(int i = h[u] ; i != -1 ; i = e[i].next){
		int v = e[i].to;
		dfs(v);
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	int flag = 0, kase = 0;
	while(scanf("%d%d", &n, &m) && (n || m)){
		if(flag) printf("\n");
		flag = 1;
		printf("Calling circles for data set %d:\n", ++kase);
		
		memset(vis, 0, sizeof(vis));
		memset(h, -1, sizeof(h));
		cnt = 0; tot = 0;
		memset(d, 0, sizeof(d));
		memset(vis, 0, sizeof(vis));
		mp.clear();
		names.clear();
		
		char s1[100], s2[100];
		
		for(int i = 1 ; i <= n ; ++i) d[i][i] = 1;
		for(int i = 1 ; i <= m ; ++i){
			scanf("%s%s", s1, s2);
			d[ID(s1)][ID(s2)] = 1;
		} 

		for(int k = 1 ; k <= n ; ++k){
			for(int i = 1 ; i <= n ; ++i){
				for(int j = 1 ; j <= n ; ++j){
					d[i][j] = d[i][j] || (d[i][k] && d[k][j]);
				}
			}
		}
		
		for(int i = 1 ; i <= n ; ++i){
			for(int j = 1 ; j <= n ; ++j){
				if(d[i][j] && d[j][i]){
					add(i, j); add(j, i);
				}
			} 
		}
		
		for(int i = 1 ; i <= n ; ++i){
			if(!vis[i]) {
				f = 0;
				dfs(i); 
				printf("\n");
			}
		}
	}
	return 0;
}

标签：Circles,闭包,cnt,ch,int,tot,Calling,maxn,mp
来源： https://www.cnblogs.com/tuchen/p/14988549.html