CF 787D Legacy（线段树思想构图+最短路）

D. Legacy

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Rick and his co-workers have made a new radioactive formula and a lot of bad guys are after them. So Rick wants to give his legacy to Morty before bad guys catch them.

There are n planets in their universe numbered from 1 to n. Rick is in planet number s (the earth) and he doesn't know where Morty is. As we all know, Rick owns a portal gun. With this gun he can open one-way portal from a planet he is in to any other planet (including that planet). But there are limits on this gun because he's still using its free trial.

By default he can not open any portal by this gun. There are q plans in the website that sells these guns. Every time you purchase a plan you can only use it once but you can purchase it again if you want to use it more.

Plans on the website have three types:

1. With a plan of this type you can open a portal from planet v to planet u.
2. With a plan of this type you can open a portal from planet v to any planet with index in range [l, r].
3. With a plan of this type you can open a portal from any planet with index in range [l, r] to planet v.

Rick doesn't known where Morty is, but Unity is going to inform him and he wants to be prepared for when he finds and start his journey immediately. So for each planet (including earth itself) he wants to know the minimum amount of money he needs to get from earth to that planet.

Input

The first line of input contains three integers n, q and s (1 ≤ n, q ≤ 105, 1 ≤ s ≤ n) — number of planets, number of plans and index of earth respectively.

The next q lines contain the plans. Each line starts with a number t, type of that plan (1 ≤ t ≤ 3). If t = 1 then it is followed by three integers v, u and w where w is the cost of that plan (1 ≤ v, u ≤ n, 1 ≤ w ≤ 109). Otherwise it is followed by four integers v, l, r and w where w is the cost of that plan (1 ≤ v ≤ n, 1 ≤ l ≤ r ≤ n, 1 ≤ w ≤ 109).

Output

In the first and only line of output print n integers separated by spaces. i-th of them should be minimum money to get from earth to i-th planet, or  - 1 if it's impossible to get to that planet.

Examples

input

3 5 1
2 3 2 3 17
2 3 2 2 16
2 2 2 3 3
3 3 1 1 12
1 3 3 17

output

0 28 12

input

4 3 1
3 4 1 3 12
2 2 3 4 10
1 2 4 16

output

0 -1 -1 12

Note

In the first sample testcase, Rick can purchase 4th plan once and then 2nd plan in order to get to get to planet number 2.

题目链接：CF 787D

很脑洞的一道题，比较好的做法是用线段树分割区间的思想，把1-n看成区间构建两颗线段树A和B，然后把所有的子区间都看成一个点，这样后面的加边就是对这些区间加边了，构建好A和B，A父子节点之间自底向上加有向边，B树自顶向下加有向边，然后把B树的子节点加无向边到对应的A树的子节点，代表这些点可以推出（到达）哪些点，由于一开始在子节点s（A[s]还是B[s]无所谓，两者本来就是强连通的），哪里都不能去，因此最后要检查的就是B树的子节点的dis值（在A数的话哪里都可以）。

然后就是三种区间操作

1：u->v，加边$<A的叶子节点u,B的叶子节点v,w>$

2：u->[l,r]，加边$<A的叶子节点u,\{B树中[l,r]区间对应的节点集合\},w>$

3：[l,r]->v，加边$<\{A树中[l,r]区间对应的节点集合\},B树的叶子节点v,w>$

这样一来三种边都不会多也不会少刚好可以到达正确的点。然后从B[s]的跑最短路即可，最后检查B[i]的dis即可，当然一开始要把A、B树的叶子节点记录下来方便后面加边使用

边数最坏情况应该是每一层左右均取两个区间，一直递归下去，有$log_{2}N$层，因此边数大约是$O(4*q*log_{2}N+2N)$

代码：

#include <stdio.h>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 1e5 + 7;
struct edge
{
    int to, nxt;
    LL w;
    edge() {}
    edge(int _to, int _nxt, LL _w): to(_to), nxt(_nxt), w(_w) {}
} E[N * 68];
int head[N << 2], tot;
LL d[N << 2];
bitset < N << 2 > vis;
int id[2][N << 2], sz;
vector<int>st;
int A[N], B[N];

void init()
{
    CLR(head, -1);
    tot = 0;
    sz = 0;
}
inline void add(int s, int t, LL w)
{
    E[tot] = edge(t, head[s], w);
    head[s] = tot++;
}
void build(int k, int l, int r, int o)
{
    id[o][k] = ++sz;
    if (l == r)
    {
        if (o)
            A[l] = id[o][k];
        else
            B[l] = id[o][k];
        return ;
    }
    else
    {
        int mid = MID(l, r);
        build(LC(k), l, mid, o);
        build(RC(k), mid + 1, r, o);
        if (o) //A树自底向上
        {
            add(id[o][LC(k)], id[o][k], 0);
            add(id[o][RC(k)], id[o][k], 0);
        }
        else//B树自顶向下
        {
            add(id[o][k], id[o][LC(k)], 0);
            add(id[o][k], id[o][RC(k)], 0);
        }
    }
}
void Findset(int k, int l, int r, int ql, int qr, int o)
{
    if (ql <= l && r <= qr)
        st.push_back(id[o][k]);
    else
    {
        int mid = MID(l, r);
        if (qr <= mid)
            Findset(LC(k), l, mid, ql, qr, o);
        else if (ql > mid)
            Findset(RC(k), mid + 1, r, ql, qr, o);
        else
            Findset(LC(k), l, mid, ql, mid, o), Findset(RC(k), mid + 1, r, mid + 1, qr, o);
    }
}
void SPFA(int s)
{
    queue<int>Q;
    Q.push(s);
    vis.reset();
    CLR(d, INF);
    d[s] = 0;
    while (!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        vis[u] = 0;
        for (int i = head[u]; ~i; i = E[i].nxt)
        {
            int v = E[i].to;
            if (d[v] > d[u] + E[i].w)
            {
                d[v] = d[u] + E[i].w;
                if (!vis[v])
                {
                    vis[v] = 1;
                    Q.push(v);
                }
            }
        }
    }
}
int main(void)
{
    int n, q, s, i, ops;
    while (~scanf("%d%d%d", &n, &q, &s))
    {
        init();
        build(1, 1, n, 1);
        build(1, 1, n, 0);
        for (i = 1; i <= n; ++i)
        {
            add(A[i], B[i], 0);
            add(B[i], A[i], 0);
        }
        for (i = 0; i < q; ++i)
        {
            scanf("%d", &ops);
            if (ops == 1)
            {
                int u, v;
                LL w;
                scanf("%d%d%I64d", &u, &v, &w);
                add(A[u], B[v], w);
            }
            else if (ops == 2)
            {
                int u, l, r;
                LL w;
                scanf("%d%d%d%I64d", &u, &l, &r, &w);
                st.clear();
                Findset(1, 1, n, l, r, 0);
                for (auto &x : st)
                    add(A[u], x, w);
            }
            else if (ops == 3)
            {
                int v, l, r;
                LL w;
                scanf("%d%d%d%I64d", &v, &l, &r, &w);
                st.clear();
                Findset(1, 1, n, l, r, 1);
                for (auto &x : st)
                    add(x, B[v], w);
            }
        }
        SPFA(B[s]);
        for (i = 1; i <= n; ++i)
            printf("%I64d%c", d[B[i]] == 0x3f3f3f3f3f3f3f3f ? -1 : d[B[i]], " \n"[i == n]);
    }
    return 0;
}