NOIP模拟（10.24）T1 建设图

建设图

题目背景：

10.24 NOIP模拟T1

分析：tarjan缩点

 

显然，本身满足双连通的部分，可以当做一个点处理，所以我们先tarjan缩一发点，然后得到了一棵树，然后我们稍加分析之后，就可以发现，只需要加入（叶子节点数 +
1） / 2条边就可以了（手动分析吧，我也不会证）

Source：

/*
created by scarlyw
*/
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cctype>
#include <set>
#include <map>
#include <vector>
#include <queue>

const int MAXN = 100000 + 10;

int n, x, y, m, o, top, ind, ans;
int low[MAXN], num[MAXN], dcc[MAXN], degree[MAXN];
std::vector<int> edge[MAXN], s, new_edge[MAXN];
bool vis[MAXN];

inline void add_edge(int x, int y) {
edge[x].push_back(y), edge[y].push_back(x);
}

inline void read_in() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= m; ++i) scanf("%d%d", &x, &y), add_edge(x, y);
}

inline void tarjan(int cur, int fa) {
low[cur] = num[cur] = ++ind, vis[cur] = true, s.push_back(cur);
for (int p = 0; p < edge[cur].size(); ++p) {
int v = edge[cur][p];
if (num[v] == 0) tarjan(v, cur), low[cur] = std::min(low[cur], low[v]);
else if (vis[v] && v != fa) low[cur] = std::min(low[cur], num[v]);
}
if (low[cur] == num[cur]) {
++top;
do {
o = s.back(), vis[o] = false, s.pop_back(), dcc[o] = top;
} while (o != cur);
}
}

inline void solve() {
tarjan(1, 1);
if (top == 1) std::cout << "0", exit(0);
for (int i = 1; i <= n; ++i)
for (int p = 0; p < edge[i].size(); ++p) {
int v = edge[i][p];
if (dcc[v] != dcc[i]) {
degree[dcc[v]]++, degree[dcc[i]]++;
}
}
int cnt = 0;
for (int i = 1; i <= top; ++i) if (degree[i] == 2) cnt++;
std::cout << (cnt + 1) / 2;
}

int main() {
//	freopen("graph.in", "r", stdin);
//	freopen("graph.out", "w", stdout);
read_in();
solve();
return 0;
}