试着写了一个 SPJ，求加入本题。

稍微自测了一下没发现出错。不知道会不会有一些边界漏判把 SPJ 搞崩掉，如果有出错麻烦告知。

此外原英文题面上：

Additionally, if the first line of your output is correct and the other two are missing or incorrect, you will receive about 70% of the value of that test case.

也就是说，如果只输出了答案不输出方案也能得到 $70\%$ 的分数，翻译可能是为了 SPJ 实现更方便加入了必须输入 5 个数的条件。我写的这个 SPJ 应该可以实现原题的标准，不知能否把计分标准换成原题的标准？

#include <testlib.h>

#include <algorithm>
#include <cstring>
#include <set>
#include <utility>

const int MaxN = 300000;

int N, Cnt;
int jans = -1, pans = -1;
int U[MaxN + 5], V[MaxN + 5];
std::set< std::pair<int, int> > Edge;

int Par[MaxN + 5];
int find(int x) { return x == Par[x] ? x : Par[x] = find(Par[x]); }

void readInput() {
  N = inf.readInt(1, MaxN, "n");
  inf.readEoln();
  for (int i = 1; i <= N; ++i) Par[i] = i;
  for (int i = 1; i < N; ++i) {
    int u = inf.readInt(1, N, "a_i");
    inf.readSpace();
    int v = inf.readInt(1, N, "b_i");
    inf.readEoln();
    ensuref(u != v, "tree can't contain loops");
    ensuref(Edge.count(std::make_pair(u, v)) == 0, "tree can't contain multiple edges between a pair of vertices");
    Edge.insert(std::make_pair(u, v));
    Edge.insert(std::make_pair(v, u));
    U[i] = std::min(u, v), V[i] = std::max(u, v);
    int p = find(u), q = find(v);
    ensuref(p != q, "tree can't contain cycles");
  }
  inf.readEof();
}

std::vector<int> Gr[MaxN + 5];
int Dis[MaxN + 5];

void dfs(int u) {
  for (int i = 0; i < (int) Gr[u].size(); ++i) {
    int v = Gr[u][i];
    if (Dis[u] + 1 < Dis[v]) {
      Dis[v] = Dis[u] + 1;
      dfs(v);
    }
  }
}

int getDiameter(int cut_u, int cut_v, int link_u, int link_v) {
  for (int i = 1; i <= N; ++i) Gr[i].clear();
  for (int i = 1; i < N; ++i)
    if (U[i] != cut_u || V[i] != cut_v) {
      Gr[U[i]].push_back(V[i]);
      Gr[V[i]].push_back(U[i]);
    }
  Gr[link_u].push_back(link_v);
  Gr[link_v].push_back(link_u);
  memset(Dis, 0x7F, sizeof Dis);
  Dis[1] = 0;
  dfs(1);
  int rt = 1;
  for (int i = 1; i <= N; ++i)
    if (Dis[i] > Dis[rt]) rt = i;
  memset(Dis, 0x7F, sizeof Dis);
  Dis[rt] = 0;
  dfs(rt);
  int dia = 0;
  for (int i = 1; i <= N; ++i)
    dia = std::max(dia, Dis[i]);
  return dia;
}

int readAns(InStream& stream) {
  int res = stream.readInt(1, N, "smallest distance between the farthest two rooms");
  stream.readEoln();
  if (stream.seekEof() == true) return res;
  int u = stream.readInt(1, N, "first room of the tunnel to be closed");
  stream.readSpace();
  int v = stream.readInt(1, N, "second room of the tunnel to be closed");
  stream.readEoln();
  int a = stream.readInt(1, N, "first room of the tunnel to be opened");
  stream.readSpace();
  int b = stream.readInt(1, N, "second room of the tunnel to be opened");
  stream.readEoln();
  stream.readEof();
  if (Edge.count(std::make_pair(u, v)) == 0) {
    if (jans == -1) quitf(_fail, "the previous tunnel doesn't exist");
    else return res;
  }
  if (u > v) std::swap(u, v);
  for (int i = 1; i <= N; ++i) Par[i] = i;
  for (int i = 1; i < N; ++i)
    if (U[i] != u || V[i] != v) Par[find(U[i])] = find(V[i]);
  int p = find(a), q = find(b);
  if (p == q) {
    if (jans == -1) quitf(_fail, "graph contains cycles after reconstruction");
    else return res;
  }
  if (getDiameter(u, v, a, b) == res) Cnt++;
  return res;
}

int main(int argc, char *argv[]) {
  registerTestlibCmd(argc, argv);
  readInput();
  jans = readAns(ans);
  pans = readAns(ouf);
  if (jans == pans) {
    if (Cnt == 2) quitf(_ok, "acceptable solution");
    else quitp(0.7, "correct answer, but bad reconstruction");
  } else if (jans < pans) {
    quitf(_wa, "jury has the better answer: jans = %d, pans = %d\n", jans, pans);
  } else {
    if (Cnt == 2) quitf(_fail, ":( participant has the better answer: jans = %d, pans = %d\n", jans, pans);
    else quitf(_wa, "participant prints a better answer, but the reconstruction can't achieve it");
  }
  return 0;
}