cf 1000E We Need More Bosses

一 原题

E. We Need More Bosses

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Your friend is developing a computer game. He has already decided how the game world should look like — it should consist of $n$ locations connected by $m$ two-way passages. The passages are designed in such a way that it should be possible to get from any location to any other location.

Of course, some passages should be guarded by the monsters (if you just can go everywhere without any difficulties, then it's not fun, right?). Some crucial passages will be guarded by really fearsome monsters, requiring the hero to prepare for battle and designing his own tactics of defeating them (commonly these kinds of monsters are called bosses). And your friend wants you to help him place these bosses.

The game will start in location $s$ and end in location $t$, but these locations are not chosen yet. After choosing these locations, your friend will place a boss in each passage such that it is impossible to get from $s$ to $t$without using this passage. Your friend wants to place as much bosses as possible (because more challenges means more fun, right?), so he asks you to help him determine the maximum possible number of bosses, considering that any location can be chosen as $s$ or as $t$.

Input

The first line contains two integers $n$ and $m$ ($2\le n\le 3\cdot {10}^5$, $n-1\le m\le 3\cdot {10}^5$) — the number of locations and passages, respectively.

Then $m$ lines follow, each containing two integers $x$ and $y$ ($1\le x,y\le n$, $x\ne y$) describing the endpoints of one of the passages.

It is guaranteed that there is no pair of locations directly connected by two or more passages, and that any location is reachable from any other location.

Output

Print one integer — the maximum number of bosses your friend can place, considering all possible choices for $s$and $t$.

Examples

input

Copy

5 5
1 2
2 3
3 1
4 1
5 2

output

Copy

2

input

Copy

4 3
1 2
4 3
3 2

output

Copy

3

二 分析

给一个联通的无向图，顶点之间没有重边。选定两个点s和t，从s到t有一些边是必经的，要求你找一对s和t，使得必经边是最多的，输出其数量。

如果输入的图没有环，那么就是求树的直径（联通且无环的无向图就是一棵树，树上两点之间的路径是唯一的）。所以把图中的环缩成一个点就可以了，如果两个环有交点，那么应当缩成一个点，使用无向图上的tarjan即可。

三 代码

/*
AUTHOR: maxkibble
LANG: c++
PROB: cf 1000E
*/

#include <cstdio>
#include <stack>
#include <vector>

const int maxn = 3e5 + 5;

#define pb push_back

int n, m;
std::vector<int> g[maxn];

void init() {
    scanf("%d%d", &n, &m);
    while (m--) {
        int s, e;
        scanf("%d%d", &s, &e);
        g[s].pb(e);
        g[e].pb(s);
    }
}

int clk, dfu[maxn], low[maxn], sz, belong[maxn];
bool inStack[maxn];
std::stack<int> st;
std::vector<std::vector<int> > comp;

void tarjan(int u, int fa) {
    dfu[u] = low[u] = ++clk;
    st.push(u);
    inStack[u] = true;
    for (int v: g[u]) {
        if (v == fa) continue;
        if (!dfu[v]) {
            tarjan(v, u);
            low[u] = std::min(low[u], low[v]);
        } else {
            low[u] = std::min(low[u], dfu[v]);
        }
    }
    if (dfu[u] == low[u]) {
        std::vector<int> v;
        sz++;
        while (true) {
            int hd = st.top();
            st.pop();
            inStack[hd] = false;
            v.pb(hd);
            belong[hd] = sz;
            if (hd == u) break;
        }
        comp.pb(v);
    }
}

std::vector<int> g1[maxn];

void buildGraph() {
    for (int i = 1; i <= n; i++) {
        for (int d: g[i]) {
            int s = belong[i], e = belong[d];
            if (s == e) continue;
            g1[s].pb(e);
        }
    }
}

int dis[maxn];

void dfs(int u, int fa) {
    dis[u] = dis[fa] + 1;
    for (int v: g1[u]) {
        if (v != fa) dfs(v, u);
    } 
}

int calcDiag() {
    dfs(1, 0);
    int furthest = 1;
    for (int i = 1; i <= sz; i++) {
        if (dis[i] > dis[furthest]) furthest = i;
    }
    dfs(furthest, 0);
    int ans = 0;
    for (int i = 1; i <= sz; i++) ans = std::max(ans, dis[i]);
    return ans - 1;
}

void solve() {
    for (int i = 1; i <= n; i++) {
        if (!dfu[i]) tarjan(i, 0);
    }
    buildGraph();
    printf("%d\n", calcDiag());
}

int main() {
    init();
    solve();
    return 0;
}