Legacy
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:
- With a plan of this type you can open a portal from planet v to planet u.
- With a plan of this type you can open a portal from planet v to any planet with index in range [l, r].
- 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 wis 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.
题意：
有n个星球，q条路径，问从一个s星球开始到其他星球的距离。
q条路径有可能有这么几种形式：
一：从一点u到v，费用为W
二：从一点u到区间[l,r]之间的任意一个星球,费用为w
三：从[l,r]之间任意一个星球到v，费用为w
思路：
不难想到直接从一个点向其他点跑出最短路就完事了。但是，观察数据我们不难发现，这样边数一下就爆炸了。考虑到是区间建边，可以采用线段树优化建图的策略，优化边数。
简单地介绍一下线段树优化建图的策略：
我们以一段1-n的区段为例子：
我们建立两棵树，一棵代表入，一棵代表入：
入树上所有的节点向他们的父亲节点连边，出树从父节点向子节点连边，所有的叶子节点之间连接上双向边（这些边的权值都是0）。这样我们可以试着推导一下，假设从v点向[l,r]之间连边，那么从v到[l,r]之间的点u就会经过v点，u所在的区间的节点再到达u，而如果从[l,r]到v连边，那么从[l,r]之间的u点就会进过u所在的区间节点到v节点，而叶子节点之间连接双向边也是保证了到达一个节点之后可以继续向其他节点走。
AC代码：
#include<cstdio>
#include<iostream>
#include<queue>
#define md(l,r) (((l)+(r))>>1)
#define rson(x) ((x)<<1|1)
#define lson(x) ((x)<<1)
typedef int Int;
#define int long long
using namespace std;
typedef long long LL;
const int size=1e5+5;
const int maxn=1e6+5;
const int inf=1e18+5;
struct node{
int l,r;
int id;
};
struct Edge{
int u,v,w;
Edge(int u=0,int v=0,int w=0):u(u),v(v),w(w){}
};
struct roar{
int id,w;
roar(int id=0,int w=0):id(id),w(w){}
friend bool operator<(roar a,roar b)
{
return a.w>b.w;
}
};
priority_queue<roar> q;
vector<Edge> edge[maxn];
int dis[maxn],vis[maxn];
int cnt=0;
void AddEdge(int u,int v,int w)
{
edge[u].push_back(Edge(u,v,w));
}
int build(int k,int l,int r,node tree[],int *no,bool type)
{
tree[k].l=l,tree[k].r=r;tree[k].id=cnt++;
if(l==r)
{
return no[l]=tree[k].id;
}
int mid=md(l,r);
if(type)//从上向下建边
AddEdge(tree[k].id,build(lson(k),l,mid,tree,no,type),0),
AddEdge(tree[k].id,build(rson(k),mid+1,r,tree,no,type),0);
else//从下向上建边
AddEdge(build(lson(k),l,mid,tree,no,type),tree[k].id,0),
AddEdge(build(rson(k),mid+1,r,tree,no,type),tree[k].id,0);
return tree[k].id;
}
void query(int k,int l,int r,int faSno,int w,node tree[],bool type)
{
if(tree[k].l>=l&&tree[k].r<=r)
{
if(type) AddEdge(faSno,tree[k].id,w);
else AddEdge(tree[k].id,faSno,w);
return ;
}
int mid=md(tree[k].l,tree[k].r);
if(r<=mid) query(lson(k),l,r,faSno,w,tree,type);
else if(l>=mid+1) query(rson(k),l,r,faSno,w,tree,type);
else query(lson(k),l,mid,faSno,w,tree,type),query(rson(k),mid+1,r,faSno,w,tree,type);
}
void Dijkstra(int beg)
{
fill(dis,dis+maxn,inf);
fill(vis,vis+maxn,inf);
while(!q.empty()) q.pop();
q.push(roar(beg,0)),dis[beg]=0;
while(!q.empty())
{
roar s=q.top();
q.pop();
int id=s.id;
if(dis[id]!=s.w) continue;
for(int i=0;i<edge[id].size();i++)
{
if(dis[edge[id][i].v]>dis[id]+edge[id][i].w)
{
dis[edge[id][i].v]=dis[id]+edge[id][i].w;
q.push(roar(edge[id][i].v,dis[edge[id][i].v]));
}
}
}
}
node intree[size<<2],outtree[size<<2];
int inno[size],outno[size];
Int main()
{
int n,q,s;
scanf("%lld%lld%lld",&n,&q,&s);
build(1,1,n,intree,inno,0);
build(1,1,n,outtree,outno,1);
for(int i=1;i<=n;i++) AddEdge(inno[i],outno[i],0),AddEdge(outno[i],inno[i],0);
for(int i=1;i<=q;i++)
{
int op;
scanf("%lld",&op);
if(op==1)
{
int u,v,w;
scanf("%lld%lld%lld",&u,&v,&w);
AddEdge(inno[u],outno[v],w);
}
else if(op==2)
{
int v,l,r,w;
scanf("%lld%lld%lld%lld",&v,&l,&r,&w);
query(1,l,r,inno[v],w,outtree,1);
}
else
{
int v,l,r,w;
scanf("%lld%lld%lld%lld",&v,&l,&r,&w);
query(1,l,r,outno[v],w,intree,0);
}
}
Dijkstra(inno[s]);
for(int i=1;i<=n;i++)
{
if(dis[outno[i]]!=inf)printf("%lld ",dis[outno[i]]);
else printf("-1 ");
}
cout<<endl;
}