洛谷 P3530 / bzoj2788【tarjan】【差分约束】

判断是否有解可以使用差分约束。

求解赛车手的成绩的取值可以使用 Floyd。但是 \(O(n^3)\) 会 TLE。

可以先进行一次缩点。

然后进行 Floyd 求出每一个连通块内的最长路径 \(long_i\)，然后最终答案是 \(\sum_{\\i=1}^{cnt} long_i\)。

存在负环就是无解。这道题无法使用 SPFA。

#include <bits/stdc++.h>

using namespace std;

/* ovo */
const int o = 0, _ = 0;
/* ovo */

const int N = 3333;
int mp[N][N];
vector <int> z[N], scc[N];
int longer[N];
int dfn[N], low[N], belong[N], counting, total, n, m1, m2;
stack <int> stk;
bool instk[N];

void doing()
{
  int p = stk.top();
  stk.pop();
  instk[p] = false;
  scc[total].push_back(p);
  belong[p] = total;
}

void dfs(int now)
{
  dfn[now] = low[now] = ++ counting;
  stk.push(now);
  instk[now] = true;
  for (auto &p : z[now])
    if (!dfn[p])
    {
      dfs(p);
      low[now] = min(low[p], low[now]);
    }
    else if (instk[p])
      low[now] = min(low[now], dfn[p]);
  if (dfn[now] == low[now])
  {
    total ++;
    while (stk.top() != now)
      doing();
    doing();
  }
}

void tarjan()
{
  for (int i = 1; i <= n; i ++)
    if (!dfn[i])
      dfs(i);
}

signed main()
{
  cin >> n >> m1 >> m2;
  memset (mp, 0x3f, sizeof mp);
  for (int i = 1; i <= n; i ++)
    mp[i][i] = 0;
  while (m1 --)
  {
    int a, b;
    cin >> a >> b;
    z[a].push_back(b);
    z[b].push_back(a);
    mp[a][b] = min(mp[a][b], -1);
    mp[b][a] = min(mp[b][a], 1);
  }
  while (m2 --)
  {
    int a, b;
    cin >> a >> b;
    z[a].push_back(b);
    mp[a][b] = min(mp[a][b], 0);
  }
  tarjan();
  for (int k = 1; k <= n; k ++)
    for (int i = 1; i <= n; i ++)
      if (belong[i] == belong[k])
        for (int j = 1; j <= n; j ++)
          if (belong[i] == belong[j])
            mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);
  for (int i = 1; i <= n; i ++)
    if (mp[i][i])
    {
      cout << "NIE\n";
      return 0;
    }
  for (int i = 1; i <= n; i ++)
    for (int j = 1; j <= n; j ++)
      if (belong[i] == belong[j])
        longer[belong[i]] = max(longer[belong[i]], mp[i][j]);
  long long result = 0;
  for (int i = 1; i <= total; i ++)
    result += longer[i] + 1;
  cout << result << '\n';
  return o^_^o;
}