大意: 給定樹, 求兩個點, 使得所有其他的點到兩點的最短距離的最大值盡量小.

二分答案轉為判定選兩個點, 向外遍歷$x$的距離是否能遍歷完整棵樹. 取直徑兩段距離$x$的位置bfs即可.

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘\n‘
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;}
//head



const int N = 2e5+100;
int n, r1, r2, ans1, ans2;
vector<int> g[N];
int vis[N], dis[N], fa[N], s[N];
vector<int> d1, d2;
queue<int> q;
int bfs(int x) {
    REP(i,0,n+10) vis[i]=0;
    REP(i,0,n+10) fa[i]=0;
    vis[x] = 1, q.push(x);
    while (!q.empty()) {
        x = q.front();q.pop();
        for (int i=0,j=g[x].size(); i<j; ++i) {
            int y = g[x][i];
            if (!vis[y]) vis[y] = 1, fa[y]=x, q.push(y);
        }
    }
    return x;
}
void init() {
    r1 = bfs(1), r2 = bfs(r1);
    *s = 0;
    for (int i=r2; i; i=fa[i]) s[++*s]=i;
}

int chk(int x) {
    ans1=d1[x],ans2=d2[x];
    REP(i,0,n+10) vis[i]=0;
    REP(i,0,n+10) dis[i]=1e9;
    dis[ans1]=dis[ans2]=0;
    q.push(ans1),q.push(ans2);
    while (q.size()) {
        int u = q.front(); q.pop();
        if (vis[u]) continue;
        vis[u] = 1;
        if (dis[u]==x) continue;
        for (int i=0,j=g[u].size(); i<j; ++i) {
            int v = g[u][i];
            if (!vis[v]) { 
                q.push(v);
                dis[v]=min(dis[v],dis[u]+1);
            }
        }
    }
    REP(i,1,n) if (!vis[i]) return 0;
    return 1;
}

void work() {
    n=rd();
    REP(i,1,n+10) g[i].clear();
    REP(i,2,n) {
        int u=rd(), v=rd();
        g[u].pb(v),g[v].pb(u);
    }
    if (n==2) return puts("0 1 2"),void();
    if (n==3) return puts("1 1 2"),void();
    init();
    if (*s<=3) return printf("1 %d %d\n",s[1],s[2]),void();
    d1.clear(),d2.clear();
    REP(i,1,*s) d1.pb(s[i]);
    PER(i,1,*s) d2.pb(s[i]);
    int l=1,r=*s-1,ans;
    while (l<=r) {
        if (chk(mid)) ans=mid,r=mid-1;
        else l=mid+1;
    }
    chk(ans);
while (ans1==ans2) { 
        if (++ans2>n) ans2=1;
    } 
    printf("%d %d %d\n",ans,ans1,ans2);
}


int main() {
    int t;
    scanf("%d", &t);
    while (t--) work();
}

Building Fire Stations ZOJ - 3820 (二分,樹的直徑)