习题7-4(uva-818)

#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int maxn = 16;
int n,matrix[maxn][maxn],vis[maxn],result,number;

int IsOpen(int base, int left) {
	return base & (1 << left);
}
//打开个数
int GetCnt(int id)
{
	int ret = 0;
	for (int i = 0; i < n; i++) {
		if (IsOpen(id,i)) ret++;
	}
	return ret;
}
//是否有3个及以上分支
int IsMoreBranch(int id)
{
	for (int i = 0; i < n; i++) {
		int cnt = 0;
		for (int j = 0; j < n; j++) {
			if(i == j || !matrix[i][j]) continue;
			//i j 没有打开
			if (!IsOpen(id,i) && !IsOpen(id, j) ) ++cnt;
			//出现3个分支
			if (cnt == 3) return true;
		}
	}
	return false;
}

bool dfs(int cur,int father, int id) {
	vis[cur] = 1;
	for (int i = 0; i < n; i++) {
		if(!matrix[cur][i]) continue;
		if (father == i || IsOpen(id, i)  ) continue;
		if (vis[i]) return true;
		if (dfs(i, cur, id)) return true;
	}
	return false;
}
//是否有环路
bool HasRing(int id)
{
	number = 0;
	memset(vis, 0, sizeof(vis));
	for (int i = 0; i < n; i++) {
		if (vis[i] || IsOpen(id,i)) continue;
		++number;
		if (dfs(i,0, id)) return true;
	}
	return false;
}

int main()
{
	int a, b, index = 0;
	while (cin >> n && n) {

		memset(matrix, 0, sizeof(matrix));
		while (cin >> a >> b && a != -1 && b != -1) {
			matrix[a-1][b-1] = matrix[b-1][a-1] = 1;
		}

		int len = 1 << n;
		result = 1e8;
		for (int i = 0; i < len; i++) {
			if(IsMoreBranch(i) || HasRing(i)) continue;
			int cnt = GetCnt(i);
			//number个集合 合并最少需要 number - 1
			if (number - 1 <= cnt) result = min(cnt, result);
		}

		cout << "Set " << ++index << ": Minimum links to open is " << result << endl;
	}
	return 0;
}