1092E Minimal Diameter Forest

解題報告

直接手玩猜貪心策略：把直徑的中點（如果有兩個，取任意一個）互相連線起來，而且一定是連成菊花圖最優。

取哪個作為菊花圖的中心點呢？繼續手玩可以猜到，一定是直徑最大的聯通塊的直徑中點。

證明也很顯然。

程式碼實現

const int MAXN = 1000 + 10;

int n, m;
std::vector<int> G[MAXN];

int cnt, bel[MAXN], maxdist[MAXN], mdnode[MAXN];
int diameter[MAXN];
int dep[MAXN][MAXN]; int fdist[MAXN];

std::vector<int> midlen;

void dfs0(int u, int fa, int b) {
    bel[u] = b;
    forall (G[u], i) {
        int v = G[u][i];
        if (v == fa) continue;
        dfs0(v, u, b);
    }
}
void dfs1(int u, int fa, int *dis) {
    forall (G[u], i) {
        int v = G[u][i];
        if (v == fa) continue;
        dis[v] = dis[u] + 1;
        dfs1(v, u, dis);
    }
}

std::vector<std::pair<int, int> > vs;

int main() {
    std::ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= m; ++i) {
        int u, v; cin >> u >> v;
        G[u].push_back(v); G[v].push_back(u);
    }
    for (int i = 1; i <= n; ++i) {
        if (!bel[i]) {
            dfs0(i, 0, ++cnt);
        } dfs1(i, 0, dep[i]);
    } memset(maxdist, 0x3f, sizeof maxdist);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (bel[i] != bel[j]) continue;
            diameter[bel[i]] = std::max(diameter[bel[i]], dep[i][j]);
        }
    } 
    for (int i = 1; i <= n; ++i) {
        int mdis = 0;
        int b = bel[i];
        for (int j = 1; j <= n; ++j) {
            if (i == j || bel[i] != bel[j]) continue;
            mdis = std::max(mdis, dep[i][j]);
        } if (mdis < maxdist[b]) {
            maxdist[b] = mdis;
            mdnode[b] = i;
        }
    } int mxdcnt = 0;
    for (int i = 1; i <= cnt; ++i) if (diameter[mxdcnt] <= diameter[i]) mxdcnt = i;
    for (int i = 1; i <= cnt; ++i) {
        if (mxdcnt == i) continue;
        G[mdnode[mxdcnt]].push_back(mdnode[i]);
        G[mdnode[i]].push_back(mdnode[mxdcnt]);
        vs.push_back({mdnode[mxdcnt], mdnode[i]});
    } dfs1(1, 0, fdist);
    int rt = 0; for (int i = 1; i <= n; ++i) if (fdist[i] > fdist[rt]) rt = i;
    memset(fdist, 0, sizeof fdist); dfs1(rt, 0, fdist);
    cout << *std::max_element(fdist + 1, fdist + 1 + n) << endl;
    for (auto v : vs) {
        cout << v.first << ' ' << v.second << endl;
    }
    return 0;
}