线段树优化建边，开两颗线段树：
对于线段树1，自顶向下连边。对于线段树2，自底向上连边。
然后对于op1我们直接连边即可。
对于op2（u -> [l, r] cost = w），这个操作在线段树1上完成即可。
对于op3（[l, r] -> v cost = w），这个操作在线段树2上完成即可。
最后跑一遍Dijkstra最短路即可。

/*Author : lifehappy
*/
#include <bits/stdc++.h>
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;const int N = 2e6 + 10;int n, m, s, tree1[N], tree2[N], tot;int head[N], to[N], nex[N], value[N], vis[N], cnt = 1;ll dis[N];void add(int x, int y, int w) {to[cnt] = y;nex[cnt] = head[x];value[cnt] = w;head[x] = cnt++;
}int build1(int rt, int l, int r) {if(l == r) {return tree1[rt] = l;}tree1[rt] = ++tot;int lid = build1(lson);int rid = build1(rson);add(tree1[rt], lid, 0);add(tree1[rt], rid, 0);return tree1[rt];
}int build2(int rt, int l, int r) {if(l == r) {return tree2[rt] = l;}tree2[rt] = ++tot;int lid = build2(lson);int rid = build2(rson);add(lid, tree2[rt], 0);add(rid, tree2[rt], 0);return tree2[rt];
}void add1(int rt, int l, int r, int u, int L, int R, int w) {if(l >= L && r <= R) {add(u, tree1[rt], w);return ;}if(L <= mid) add1(lson, u, L, R, w);if(R > mid)  add1(rson, u, L, R, w);
}void add2(int rt, int l, int r, int v, int L, int R, int w) {if(l >= L && r <= R) {add(tree2[rt], v, w);return ;}if(L <= mid) add2(lson, v, L, R, w);if(R > mid)  add2(rson, v, L, R, w);
}typedef pair<ll, int> pli;priority_queue<pli, vector<pli>, greater<pli> > q;void Dijkstra() {memset(dis, 0x3f, sizeof dis);dis[s] = 0;q.push(make_pair(0, s));while(q.size()) {int temp = q.top().second;q.pop();if(vis[temp]) continue;vis[temp] = 1;for(int i = head[temp]; i; i = nex[i]) {if(dis[to[i]] > dis[temp] + value[i]) {dis[to[i]] = dis[temp] + value[i];q.push(make_pair(dis[to[i]], to[i]));}   }}for(int i = 1; i <= n; i++) {printf("%lld%c", dis[i] == 0x3f3f3f3f3f3f3f3f ? -1 : dis[i], i == n ? '\n' : ' ');}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d %d %d", &n, &m, &s);tot = n;build1(1, 1, n);build2(1, 1, n);for(int i = 1; i <= m; i++) {int op, x, y, l, r, w;scanf("%d", &op);if(op == 1) {scanf("%d %d %d", &x, &y, &w);add(x, y, w);}else if(op == 2) {scanf("%d %d %d %d", &x, &l, &r, &w);//x -> [l, r] cost w.add1(1, 1, n, x, l, r, w);}else {scanf("%d %d %d %d", &x, &l, &r, &w);//[l, r] -> x cost w.add2(1, 1, n, x, l, r, w);}}Dijkstra();return 0;
}