codeforce 1000 E. We Need More Bosses

最新推荐文章于 2023-05-08 19:49:22 发布

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本文介绍了一种在游戏世界中优化Boss布局的方法，通过强连通分量缩点找到树的直径来确定最大数量的Boss位置，确保从起点到终点路径上的挑战最大化。

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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 $m$ ( $2 \leq n \leq 3 \cdot 10^{5}$ , $n - 1 \leq m \leq 3 \cdot 10^{5}$ ) — the number of locations and passages, respectively.

Then $m$ lines follow, each containing two integers $x$ and $y$ ( $1 \leq x, y \leq n$ , $x \neq y$ ) 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 $t$

思路：
强连通分量缩点，找树的直径。
代码：
#include<bits/stdc++.h>
using namespace std;
const int maxn=3e5+10;
vector<int>G[maxn],mp[maxn];
int dfn[maxn],low[maxn],fa[maxn],d[maxn];
int tol=0,top=0;
bool vis[maxn];
stack<int>P;
void tarjan(int v,int Fa)
{
    dfn[v]=low[v]=++tol;
    P.push(v);
    for(int i=0;i<G[v].size();i++)
    {
        int to=G[v][i];
        if(to==Fa)continue;
        if(!dfn[to])
        {
            tarjan(to,v);
            low[v]=min(low[v],low[to]);
        }
        else low[v]=min(low[v],dfn[to]);
    }
    if(low[v]==dfn[v])
    {
        top++;int s;
        do
        {
            s=P.top();P.pop();
            fa[s]=top;
        }
        while(s!=v);
    }
}
void bfs(int st)
{
    memset(vis,0,sizeof(vis));
    queue<int>P;
    P.push(st);vis[st]=1; d[st]=0;
    while(!P.empty())
    {
        int v=P.front();P.pop();
        for(int i=0;i<mp[v].size();i++)
        {
            int to=mp[v][i];
            if(vis[to])continue;
            d[to]=d[v]+1;
            P.push(to);
            vis[to]=1;
        }
    }
}
int main()
{
    int n,m;scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int x,y;scanf("%d%d",&x,&y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    tarjan(1,-1);
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<G[i].size();j++)
        {
            int to=G[i][j];
            if(fa[i]!=fa[to])
                mp[fa[i]].push_back(fa[to]);
        }
    }
    bfs(1);
    int id=1;
    for(int i=2;i<=top;i++)
        if(d[i]>d[id])id=i;
    bfs(id);
    int ans=0;
    for(int i=1;i<=top;i++)
        ans=max(ans,d[i]);
    printf("%d\n",ans);
    return 0;
}