[CF786B]Legacy

Legacy

Subject to the effect

Maintain a map, you can at any time:

1, an edge is added \ ((u, v) \ )

2, a set of edges added \ ((u, {l ~ r}) \)

3, a set of edges added \ (({l ~ r} , u) \)

All sides are set above the weights, from a certain point and finally find the shortest path to all points.

Solution

If violence will be built side + dij T's

So we consider the tree line side by Optimized

We have established two tree line, a one, not even to his father side, the other side from even a b son from father to son, and then leaves b corresponds to a leaf edges even, uniform weights are 0

Then if a single point to single point even edges, it leaves from a leaf connected to the b side is connected

Even if the single-point interval, it is even interval from the side of b to a son

Point is connected to a single section from a section connected to the side of the son b

The last run again dij enough.

In addition, SPFA will be T

code:

#include<bits/stdc++.h>
using namespace std;
#define int long long
struct qwq{
    int v,w;
    int nxt;
}edge[4000010];
int cnt=-1;
int head[1000010];
void add(int u,int v,int w){
    edge[++cnt].nxt=head[u];
    edge[cnt].v=v;
    edge[cnt].w=w;
    head[u]=cnt;
}
int numa[1000010],numb[1000010];
int re1[1000010],re2[1000010];
int tot;
void builda(int o,int l,int r){
    numa[o]=++tot;
    if(l==r){
        re1[l]=tot;
        return;
    }
    int mid=(l+r)/2;
    builda(o*2,l,mid);
    builda(o*2+1,mid+1,r);
    add(numa[o*2],numa[o],0);
    add(numa[o*2+1],numa[o],0);
}
void buildb(int o,int l,int r){
    numb[o]=++tot;
    if(l==r){
        re2[l]=tot;
        add(re2[l],re1[l],0);
        return;
    }
    int mid=(l+r)/2;
    buildb(o*2,l,mid);
    buildb(o*2+1,mid+1,r);
    add(numb[o],numb[o*2],0);
    add(numb[o],numb[o*2+1],0);
}
void update1(int o,int l,int r,int L,int R,int v,int w){
    if(L<=l&&r<=R){
        add(numa[o],v,w);
        return;
    }
    int mid=(l+r)/2;
    if(L<=mid)update1(o*2,l,mid,L,R,v,w);
    if(mid<R)update1(o*2+1,mid+1,r,L,R,v,w);
}
void update2(int o,int l,int r,int L,int R,int v,int w){
    if(L<=l&&r<=R){
        add(v,numb[o],w);
        return;
    }
    int mid=(l+r)/2;
    if(L<=mid)update2(o*2,l,mid,L,R,v,w);
    if(mid<R)update2(o*2+1,mid+1,r,L,R,v,w);
}
int dis[1000010];
struct QAQ{
    int v,w;
    bool operator <(const QAQ& tmp)const{
        return w>tmp.w;
    }
};
void dijkstra(int s){
    for(int i=1;i<=1000000;++i)dis[i]=100000000000000000ll;
    dis[s]=0;
    priority_queue<QAQ> q;
    q.push((QAQ){s,0});
    while(!q.empty()){
        QAQ u=q.top();
        q.pop();
        int v=u.v,w=u.w;
        if(w!=dis[v])continue;
        for(int i=head[v];~i;i=edge[i].nxt){
            int tv=edge[i].v,tw=edge[i].w;
            if(dis[tv]>dis[v]+tw){
                dis[tv]=dis[v]+tw;
                q.push((QAQ){tv,dis[tv]});
            }
        }
    }
}
signed main(){
    memset(head,-1,sizeof(head));
    int n,q,s;
    scanf("%lld%lld%lld",&n,&q,&s);
    builda(1,1,n);
    buildb(1,1,n);
    for(int i=1;i<=q;++i){
        int opt;
        scanf("%d",&opt);
        if(opt==1){
            int u,v,w;
            scanf("%lld%lld%lld",&u,&v,&w);
            add(re1[u],re2[v],w);
        }
        else if(opt==2){
            int u,l,r,w;
            scanf("%lld%lld%lld%lld",&u,&l,&r,&w);
            update2(1,1,n,l,r,re1[u],w);
        }
        else {
            int u,l,r,w;
            scanf("%lld%lld%lld%lld",&u,&l,&r,&w);
            update1(1,1,n,l,r,re2[u],w);
        }
    }
    dijkstra(re1[s]);
    for(int i=1;i<=n;++i){
        printf("%lld ",dis[re1[i]]>10000000000000ll?-1:dis[re1[i]]);
    }
}