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2 years ago
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int N = 100100 << 3, M = N << 2;
typedef pair<LL, int> PII;
//最短路径
LL d[N]; //最短距离数组
bool st[N];
//邻接表
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
struct Node {
int l, r;
int ls, rs; //左儿子编号,右儿子编号
} tr[N];
int rtin, rtout, cnt; //出树根编号,入树根编号,编号计数器(从n+1开始前面n个留给真实节点)
// l:左边界, r:右边界, 返回新创建的节点编号
int build1(int l, int r) {
if (l == r) {
tr[l] = {l, r}; // tr[l]:叶子节点比较牛B不用计数器生成节点号而是真实的[1~n]。这样才能真正实现最底层的重复利用,防止重复建设
return l;
}
int u = ++cnt; //非叶子节点利用cnt获取节点号
tr[u] = {l, r}; //记录区间范围
int mid = (l + r) >> 1;
tr[u].ls = build1(l, mid); //构建左子树
tr[u].rs = build1(mid + 1, r); //构建右子树
//父节点向左右儿子连权值为0的边
add(u, tr[u].ls, 0), add(u, tr[u].rs, 0);
//返回新创建的节点编号
return u;
}
int build2(int l, int r) {
if (l == r) {
tr[l] = {l, r};
return l;
}
int u = ++cnt;
tr[u] = {l, r};
int mid = (l + r) >> 1;
tr[u].ls = build2(l, mid);
tr[u].rs = build2(mid + 1, r);
//左右儿子向父节点连权值为0的边
add(tr[u].ls, u, 0), add(tr[u].rs, u, 0);
return u;
}
/**
* @brief
*
* @param u 线u
* @param x
* @param l
* @param r
* @param w
*/
void add1(int u, int x, int l, int r, int w) {
if (l <= tr[u].l && tr[u].r <= r) {
add(x, u, w);
return;
}
if (l <= tr[tr[u].ls].r) add1(tr[u].ls, x, l, r, w);
if (r >= tr[tr[u].rs].l) add1(tr[u].rs, x, l, r, w);
}
void add2(int u, int x, int l, int r, int w) {
if (l <= tr[u].l && tr[u].r <= r) {
add(u, x, w);
return;
}
if (l <= tr[tr[u].ls].r) add2(tr[u].ls, x, l, r, w);
if (r >= tr[tr[u].rs].l) add2(tr[u].rs, x, l, r, w);
}
void dijkstra(int s) {
memset(d, 0x3f, sizeof(d));
d[s] = 0;
priority_queue<PII, vector<PII>, greater<PII>> q;
q.push({0, s});
while (q.size()) {
int u = q.top().second;
q.pop();
if (st[u]) continue;
st[u] = true;
for (int i = h[u]; ~i; i = ne[i]) {
int j = e[i];
if (d[j] > d[u] + w[i]) {
d[j] = d[u] + w[i];
q.push({d[j], j});
}
}
}
}
int main() {
//加快读入
ios::sync_with_stdio(false), cin.tie(0);
memset(h, -1, sizeof h);
int n, q, s;
cin >> n >> q >> s;
cnt = n; //可用的合法编号从n+1开始
rtin = build1(1, n); //构建入树
rtout = build2(1, n); //构建出树
int op, x, y, l, r, w;
for (int i = 1; i <= q; i++) {
cin >> op;
if (op == 1) { //点x到点y有一条边权w的边
cin >> x >> y >> w;
add(x, y, w);
} else if (op == 2) { //点x到区间[l,r]有一条边权w的边
cin >> x >> l >> r >> w;
add1(rtin, x, l, r, w);
} else if (op == 3) { //区间[l,r]到点x有一条边权w的边
cin >> x >> l >> r >> w;
add2(rtout, x, l, r, w);
}
}
//最短路径
dijkstra(s);
//输出最短路
for (int i = 1; i <= n; i++)
printf("%lld ", d[i] == INF ? -1 : d[i]);
putchar('\n');
return 0;
}