Codeforces 1092E Minimal Diameter Forest

Minimal Diameter Forest

首先我们找出每个连通块中的特殊点， 特殊点的定义是到各种个连通块中距离的最大值最小的点， 

每个连通块肯定通过特殊点连到其他连通块， 我们把有最大值的特殊点当作根， 然后其他点直接接在这个点中， 形成菊花图。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 1000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, m, cnt, mx, DIA, belong[N], mn[N], who[N], id[N], dia[N];
vector<int> G[N];

void dfs(int u) {
    belong[u] = cnt;
    for(auto& v : G[u])
        if(!belong[v]) dfs(v);
}
void dfs2(int u, int fa, int depth) {
    mx = max(mx, depth);
    for(auto& v : G[u])
        if(v != fa) dfs2(v, u, depth + 1);
}

bool cmp(const int& x, const int& y) {
    return mn[x] > mn[y];
}

int main() {
    memset(mn, inf, sizeof(mn));
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++) {
        int u, v; scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    for(int i = 1; i <= n; i++) {
        if(!belong[i]) ++cnt, dfs(i);
    }
    for(int i = 1; i <= n; i++) {
        mx = 0;
        dfs2(i, 0, 0);
        if(mx < mn[belong[i]]) {
            mn[belong[i]] = mx;
            who[belong[i]] = i;
        }
        dia[belong[i]] = max(dia[belong[i]], mx);
    }
    for(int i = 1; i <= cnt; i++) {
        id[i] = i;
        DIA = max(DIA, dia[i]);
    }
    sort(id + 1, id + 1 + cnt, cmp);
    if(cnt >= 2) DIA = max(DIA, mn[id[1]] + mn[id[2]] + 1);
    if(cnt >= 3) DIA = max(DIA, mn[id[2]] + mn[id[3]] + 2);
    printf("%d\n", DIA);
    for(int i = 2; i <= cnt; i++) printf("%d %d\n", who[id[1]], who[id[i]]);
    return 0;
}

/*
*/