#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ri register
#define pii pair<int,int>
const int inf = 0x3f3f3f3f;
inline int rd() {
int x = 0, y = 1;
char c = getchar();
for (; c < '0' || c > '9'; c = getchar())
if (c == '-')
y = -1;
for (; c >= '0' && c <= '9'; c = getchar())
x = (x << 1) + (x << 3) + (c ^ 48);
return x * y;
}
const int N = 1e4 + 5;
int siz[N], n, m, u, v, cnt, num[N], bar[N], dfn[N], low[N], tot, ans[N];
vector<int>e[N];
void dfs(int x) {
num[x] = cnt;
bar[cnt]++;
for (auto i : e[x])
if (!num[i])
dfs(i);
}
void tarjan(int x) {
dfn[x] = low[x] = ++tot;
siz[x] = 1;
int r = bar[num[x]] - 1;
for (auto i : e[x])
if (dfn[i])
low[x] = min(low[x], dfn[i]);
else {
tarjan(i);
siz[x] += siz[i];
low[x] = min(low[x], low[i]);
if (low[i] >= dfn[x])
ans[x]++, r -= siz[i];
}
ans[x] += r > 0;
}
signed main() {
while (1) {
cnt = tot = 0;
memset(num, 0, sizeof num);
memset(bar, 0, sizeof bar);
memset(low, 0, sizeof low);
memset(dfn, 0, sizeof dfn);
memset(siz, 0, sizeof siz);
memset(ans, 0, sizeof ans);
n = rd();
m = rd();
for (int i = 1; i <= n; i++)
e[i].clear();
if (!n && !m)
break;
while (m--)
u = rd() + 1, v = rd() + 1, e[u].push_back(v), e[v].push_back(u);
for (int i = 1; i <= n; i++)
if (!num[i])
++cnt, dfs(i);
for (int i = 1; i <= n; i++)
if (!dfn[i])
tarjan(i);
int s = 0;
for (int i = 1; i <= n; i++)
s = max(s, ans[i] + cnt - 1);
cout << s << '\n';
}
return 0;
}
一本通1525:电力
最新推荐文章于 2025-12-16 20:57:28 发布
博客围绕图论和算法展开,运用C++语言进行相关实现。图论在信息技术领域有重要应用,结合C++语言能高效处理相关问题,可用于解决多种实际场景中的算法需求。
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