题目大意:给一张无向图,要求找一对$s$和$t$,使得其路径上的割边是最多的,输出其数量。

题解:把边双缩点以后求树的直径。

卡点:

C++ Code:

#include <cstdio>
#include <cstring>
#define maxn 300010
#define maxm 300010 #define ONLINE_JUDGE
#define read() R::READ()
#include <cctype>
namespace R {
int x;
#ifdef ONLINE_JUDGE
char *ch, op[1 << 26];
inline void init() {
fread(ch = op, 1, 1 << 26, stdin);
}
inline int READ() {
while (isspace(*ch)) ch++;
for (x = *ch & 15, ch++; isdigit(*ch); ch++) x = x * 10 + (*ch & 15);
return x;
}
#else
char ch;
inline int READ() {
ch = getchar();
while (isspace(ch)) ch = getchar();
for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
return x;
}
#endif
} int n, m;
inline int min(int a, int b) {return a < b ? a : b;} namespace Tree {
int head[maxn], cnt;
struct Edge {
int to, nxt;
} e[maxm << 1];
void add(int a, int b) {
e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
e[++cnt] = (Edge) {a, head[b]}; head[b] = cnt;
} int q[maxn], h = 1, t = 0;
int d[maxn];
int bfs(int rt) {
memset(d, 0, sizeof d);
d[q[h = t = 0] = rt] = 1;
while (h <= t) {
int u = q[h++];
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (!d[v]) {
d[v] = d[u] + 1;
q[++t] = v;
}
}
}
return q[t];
}
int work() {
int x = bfs(1), y = bfs(x);
return d[y] - 1;
}
}
namespace Graph {
int head[maxn], cnt;
struct Edge {
int from, to, nxt;
} e[maxm << 1];
void add(int a, int b) {
e[++cnt] = (Edge) {a, b, head[a]}; head[a] = cnt;
e[++cnt] = (Edge) {b, a, head[b]}; head[b] = cnt;
} int res[maxn], CNT;
int DFN[maxn], low[maxn], idx, fa[maxn];
int S[maxn], top;
void Tarjan(int u) {
DFN[u] = low[u] = ++idx;
S[++top] = u;
int v;
for (int i = head[u]; i; i = e[i].nxt) {
v = e[i].to;
if (v == fa[u]) continue;
if (!DFN[v]) {
fa[v] = u;
Tarjan(v);
low[u] = min(low[u], low[v]);
if (low[v] > DFN[u]) {
CNT++;
do res[v = S[top--]] = CNT; while (v != e[i].to);
}
} else low[u] = min(low[u], DFN[v]);
}
}
inline void tarjan(int n) {
for (int i = 1; i <= n; i++) if (!DFN[i]) Tarjan(i);
}
inline void init() {
for (int i = 1; i <= cnt; i += 2) {
if (res[e[i].from] != res[e[i].to]) Tree::add(res[e[i].from], res[e[i].to]);
}
}
} int main() {
#ifdef ONLINE_JUDGE
R::init();
#endif
n = read(), m = read();
for (int i = 1; i <= m; i++) Graph::add(read(), read());
Graph::tarjan(n);
Graph::init();
printf("%d\n", Tree::work());
return 0;
}

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