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##[$Legacy$](https://codeforces.com/contest/787/problem/D)
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[洛谷题目传送门](https://www.luogu.com.cn/problem/CF786B)
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[视频讲解](https://www.bilibili.com/video/BV1Tk4y1m7VM?p=12)
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### 一、题目大意
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### 二、解决思路
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<font color='red' size=5><b>线段树优化建图的板子题</b></font>
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考虑暴力建图。显然不能通过此题。$1≤n,q≤10^5,1≤w≤10^9$,数量太大！
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这时候就需要用线段树优化建图了。线段树优化建图就是 **利用线段树**，**减少连边数量**，从而降低复杂度。
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#### 基本思想
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先建一棵线段树。假如现在我们要从 $8$ 号点向区间 $[3,7]$ 的所有点连一条权值为 $w$ 有向边。
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<center><img src='https://img2020.cnblogs.com/blog/1859218/202010/1859218-20201003152331500-2047892066.png'></center>
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那么怎么连边？把区间 $[3,7]$ 拆成 $[3,4]$、$[5,6]$ 和 $[7,7]$ 然后分别连边。
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就这样：（如下图所示。其中黑色普通边的边权为 $0$，粉色边的边权为 $w$。）
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<center><img src='https://img2020.cnblogs.com/blog/1859218/202010/1859218-20201003152713737-813466449.png'></center>
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原来我们要连 $5$ 条边，现在只需要连 $3$ 条边，也就是 $⌈log_27⌉$ 条边。
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于是 $O(n)$ 的边数就优化成了 $O(logn)$。
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那么 **操作三 ($op=3$)** 用和 **操作二 ($op=2$)** 类似的方法连边。从区间 $[3,7]$ 的所有点向 $8$ 号点连一条权值为 $w$ 有向边：（其实就是边反了个方向）
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<center><img src='https://img2020.cnblogs.com/blog/1859218/202010/1859218-20201003155545073-505477788.png'></center>
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以上是操作二与操作三分开来考虑的情形，那么操作二与操作三相结合该怎么办呢？
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考虑建两棵线段树，第一棵只连自上而下的边，第二棵只连自下而上的边。方便起见，我们把第一棵树称作“**出树**”，第二棵树称作“**入树**”。
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初始时自上而下或自下而上地在每个节点与它的父亲之间连边。由于两棵线段树的叶子节点实际上是同一个点，因此要在它们互相之间连边权为 $0$ 的边。初始时是这样的：
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<center><img src='https://img2020.cnblogs.com/blog/1859218/202010/1859218-20201003175442575-1059055629.png'></center>
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接下来：
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* 对于操作一，就从入树的叶子节点向出树的叶子节点连边。
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* 对于操作二，就从入树的叶子节点向出树中的对应区间连边。
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* 对于操作三，就从入树中的对应区间向出树中的叶子节点连边。
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举个栗子。比如现在我们要从 $8$ 号点向区间 $[3,7]$ 的所有点连一条权值为 $w$ 有向边。那么就如图所示连边：（为了让图更清楚，图中把入树和出树叶子节点之间相连的边省略了。）
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<center><img src='https://img2020.cnblogs.com/blog/1859218/202010/1859218-20201003180159361-1510386633.png'></center>
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### 三、代码实现
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const LL INF = 0x3f3f3f3f3f3f3f3f;
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const int N = 100100 << 3, M = N << 2;
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typedef pair<LL, int> PII;
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//最短路径
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LL d[N]; //最短距离数组
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bool st[N];
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//邻接表
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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struct Node {
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int l, r;
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int ls, rs; //左儿子编号，右儿子编号
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} tr[N];
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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)
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// l:左边界, r:右边界, 返回新创建的节点编号
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int build1(int l, int r) {
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if (l == r) {
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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
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return l;
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}
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int u = ++cnt; //非叶子节点利用cnt获取节点号
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tr[u] = {l, r}; //记录区间范围
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int mid = (l + r) >> 1;
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tr[u].ls = build1(l, mid); //构建左子树
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tr[u].rs = build1(mid + 1, r); //构建右子树
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//父节点向左右儿子连权值为0的边
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add(u, tr[u].ls, 0), add(u, tr[u].rs, 0);
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//返回新创建的节点编号
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return u;
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}
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int build2(int l, int r) {
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if (l == r) {
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tr[l] = {l, r};
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return l;
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}
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int u = ++cnt;
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tr[u] = {l, r};
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int mid = (l + r) >> 1;
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tr[u].ls = build2(l, mid);
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tr[u].rs = build2(mid + 1, r);
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//左右儿子向父节点连权值为0的边
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add(tr[u].ls, u, 0), add(tr[u].rs, u, 0);
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return u;
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}
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/**
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* @brief 点向区间连边 或 区间向点连边
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*
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* @param u 线段树根节点u
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* @param x 需要连边的点
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* @param l 区间左边界
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* @param r 区间右边界
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* @param w 权值
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*/
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void add1(int u, int x, int l, int r, int w) {
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if (l <= tr[u].l && tr[u].r <= r) {
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add(x, u, w);
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return;
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}
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if (l <= tr[tr[u].ls].r) add1(tr[u].ls, x, l, r, w);
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if (r >= tr[tr[u].rs].l) add1(tr[u].rs, x, l, r, w);
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}
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void add2(int u, int x, int l, int r, int w) {
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if (l <= tr[u].l && tr[u].r <= r) {
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add(u, x, w);
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return;
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}
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if (l <= tr[tr[u].ls].r) add2(tr[u].ls, x, l, r, w);
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if (r >= tr[tr[u].rs].l) add2(tr[u].rs, x, l, r, w);
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}
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void dijkstra(int s) {
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memset(d, 0x3f, sizeof(d));
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d[s] = 0;
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priority_queue<PII, vector<PII>, greater<PII>> q;
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q.push({0, s});
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while (q.size()) {
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int u = q.top().second;
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q.pop();
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if (st[u]) continue;
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st[u] = true;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (d[j] > d[u] + w[i]) {
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d[j] = d[u] + w[i];
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q.push({d[j], j});
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}
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}
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}
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}
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int main() {
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//加快读入
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ios::sync_with_stdio(false), cin.tie(0);
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memset(h, -1, sizeof h);
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int n, q, s;
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cin >> n >> q >> s;
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cnt = n; //可用的合法编号从n+1开始
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rtin = build1(1, n); //构建入树
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rtout = build2(1, n); //构建出树
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int op, x, y, l, r, w;
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for (int i = 1; i <= q; i++) {
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cin >> op;
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if (op == 1) { //点x到点y有一条边权w的边
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cin >> x >> y >> w;
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add(x, y, w);
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} else if (op == 2) { //点x到区间[l,r]有一条边权w的边
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cin >> x >> l >> r >> w;
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add1(rtin, x, l, r, w);
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} else if (op == 3) { //区间[l,r]到点x有一条边权w的边
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cin >> x >> l >> r >> w;
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add2(rtout, x, l, r, w);
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}
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}
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//最短路径
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dijkstra(s);
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//输出最短路
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for (int i = 1; i <= n; i++)
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printf("%lld ", d[i] == INF ? -1 : d[i]);
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putchar('\n');
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return 0;
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}
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```
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