$Dijkstra$ 模板，求解单源最短路，复杂度 $O(m \log(n))$，其中 $n$ 为点数，$m$ 为边数。

思想是把已经求出最短距离的点加入到集合中，然后把其它点到起点的距离加入到最小堆中，每次从堆顶取出元素 $i$，用它去做松弛操作，如果能够更新相邻点 $j$，那么把 $j$ 的距离加入到最小堆中。可以保证每次从堆顶取出元素之后，这个元素的最短距离就已经被确定，可以直接把它加入到集合中。

算法要求边权值都是正数。

// Date: Fri Dec 29 11:33:23 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nemp(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

template <typename t> istream &operator>>(istream &in, vector<t> &vec) {
  for (t &x : vec)
    in >> x;
  return in;
}

template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) {
  int n = SZ(vec);
  For(i, 0, n) {
    out << vec[i];
    if (i < n - 1)
      out << ' ';
  }
  return out;
}

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 100100, M = 200100;

int n, m, s, h[N], st[N], dis[N];
int idx, e[M], ne[M], w[M];

void Init() {
  idx = 0;
  memset(h, -1, sizeof h);
  memset(st, 0, sizeof st);
  memset(dis, 0x3f, sizeof dis);
}

void Add(int a, int b, int c) {
  e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

void dijkstra(int s) {
  dis[s] = 0;

  priority_queue<PII, vector<PII>, greater<PII>> q;
  q.push({0, s});

  while (nemp(q)) {
    auto t = q.top();
    q.pop();

    int base = t.f1, ver = t.f2;
    if (st[ver])
      continue;
    st[ver] = true;

    ForE(i, ver) {
      int j = e[i], tmp = base + w[i];
      if (tmp < dis[j]) {
        dis[j] = tmp;
        q.push({tmp, j});
      }
    }
  }
}

int main(void) {
#ifdef _DEBUG
  freopen("4779.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> n >> m >> s;
  Init();
  For1(i, 1, m) {
    int a, b, c;
    cin >> a >> b >> c;
    Add(a, b, c);
  }
  dijkstra(s);

  For1(i, 1, n) {
    int tmp = dis[i];
    if (tmp == INF)
      cout << ((1LL << 31) - 1) << ' ';
    else
      cout << tmp << ' ';
  }
  cout << '\n';

  return 0;
}