zoj 1119 SPF

利用并查集做

设某个要计算的点为p
然后对与p连接的那些点做深度搜索，然后在dfs中对他们经行合并操作。
最后判断祖先的数量有多少，若>1，则有几个祖先就说明有几个不一样的子集，也就是不能互相访问的子网。

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

vector map[1001];
int anc[1001];
int maxn = 0;
int vis[1001];
int val[1001];
int temp[1001];
void init() {
	for (int i = 0; i <= maxn; i++) {
		map[i].clear();
	}
	maxn = 0;
}


int Find(int id) {
	return (anc[id] == id) ? id : (anc[id] = Find(anc[id]));
}

void unio(int p1, int p2) {
	int fp1 = Find(p1);
	int fp2 = Find(p2);

	if (val[fp1] > val[fp2]) {
		anc[p2] = fp1;
		val[fp1] += val[fp2];
	} else {
		anc[p1] = fp2;
		val[fp2] += val[fp1];
	}
}


void dfs(int id) {
	vector nd = map[id];
	for (int i = 0; i < nd.size(); i++) {
		if (!vis[nd[i]]) {
			vis[nd[i]] = 1;
			unio(id, nd[i]);
			dfs(nd[i]);
		}
	}
}

int main() {
	int u, v, cur;
	cur = 0;
	while (cin >> u) {
		if (!u) break;
		if(cur){
			cout << endl;
		}
		cin >> v;
		init();

		map[u].push_back(v);
		map[v].push_back(u);
		maxn = max(maxn, max(u, v));
		while (cin >> u) {
			if (!u) break;
			cin >> v;
			map[u].push_back(v);
			map[v].push_back(u);
			maxn = max(maxn, max(u, v));
		}



		printf("Network #%dn", ++cur);
		int flag=0;
		for (int i = 1; i <= maxn; i++) {
			vector roots = map[i];
			memset(vis, 0, sizeof(vis));
			memset(temp, 0, sizeof(temp));
			
			//初始化
			for (int i = 1; i <= maxn; i++) {
				anc[i] = i;
				val[i] = 1;
			}
			vis[i] = 1;
			for (int j = 0; j < roots.size(); j++) {
				if (vis[roots[j]])continue;
				vis[roots[j]] = 1;
				dfs(roots[j]);
			}


			int res = 0;
			for (int j = 0; j < roots.size(); j++) {
				int fa = Find(roots[j]);
				if (!temp[fa]) {
					res++;
					temp[fa] = 1;
				}
			}
			
//			printf("forbid i:%dn", i);
//			for (int j = 1; j <= maxn; j++) {
//				cout << j << "anc:" << anc[j] << endl;
//			}

			if (res > 1) {
				printf("  SPF node %d leaves %d subnetsn", i, res);
				flag=1;
			}
		}
		if(!flag){
			cout << "  No SPF nodes"<