uva 10972 - RevolC FaeLoN（双联通）

题目链接：uva 10972 - RevolC FaeLoN

将图缩点，每个双联通分量中的两点一定可以相互到达。缩完点之后就是一棵树，也有可能是森林，需要建的变数即为整个森林的叶子节点个数除2，需要注意的是当一个树只有一个节点的时候，需要和其他节点建一条入边，一条出边，贡献度为2.

#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <algorithm>

using namespace std;
typedef pair<int,int> pii;
const int maxn = 1005;
const int maxm = 1e6;

int N, M, E, first[maxn], jump[maxm * 2], link[maxm * 2], iscut[maxm * 2];
int cntlock, cntbcc, pre[maxn], bccno[maxn];
vector<int> G[maxn], BCC[maxn];

int dfs (int u, int fa) {
int lowu = pre[u] = ++cntlock;

for (int i = first[u]; i != -1; i = jump[i]) {
int v = link[i];

if (!pre[v]) {
int lowv = dfs(v, u);
lowu = min(lowu, lowv);
if (lowv > pre[u])
iscut[i] = iscut[i^1] = 1;
} else if (pre[v] < pre[u] && v != fa)
lowu = min(lowu, pre[v]);
}
return lowu;
}

void dfs (int u) {
bccno[u] = cntbcc;
BCC[cntbcc].push_back(u);

for (int i = first[u]; i != -1; i = jump[i]) {
if (iscut[i]) continue;
int v = link[i];
if (!bccno[v]) dfs(v);
}
}

void findBCC() {
cntlock = cntbcc = 0;
memset(pre, 0, sizeof(pre));
memset(iscut, 0, sizeof(iscut));
memset(bccno, 0, sizeof(bccno));
for (int i = 1; i <= N; i++)
if (!pre[i]) dfs(i, -1);

for (int i = 1; i <= N; i++) {
if (!bccno[i]) {
BCC[++cntbcc].clear();
dfs(i);
}
}
}

inline void addEdge(int u, int v) {
jump[E] = first[u];
link[E] = v;
first[u] = E++;
}

void init () {
E = 0;
memset(first, -1, sizeof(first));

int u, v;
while (M--) {
scanf("%d%d", &u, &v);
addEdge(u, v);
addEdge(v, u);
}
findBCC();
}

void search(int u, int& c) {
pre[u] = 1;
if (G[u].size() <= 1) c++;
for (int i = 0; i < G[u].size(); i++)
if (!pre[G[u][i]]) search(G[u][i], c);
}

int main () {
while (scanf("%d%d", &N, &M) == 2) {
init();

for (int i = 1; i <= cntbcc; i++) G[i].clear();
for (int i = 1; i <= N; i++) {
for (int j = first[i]; j != -1; j = jump[j]) {
if (iscut[j]) {
int u = bccno[i], v = bccno[link[j]];
G[u].push_back(v);
}
}
}

int ans = 0;
memset(pre, 0, sizeof(pre));
for (int i = 1; i <= cntbcc; i++) {
if (!pre[i]) {
int cnt = 0;
search(i, cnt);
ans += (cnt == 1 ? 2 : cnt);
}
}

if (cntbcc == 1) printf("0\n");
else printf("%d\n", (ans+1) / 2);
}
return 0;
}