目录

20190809 review——时之终结

20190809 review——时之终结

思路

考虑二进制拆分,建出$log_2^Y$的满图,如果$Y$二进制拆分后第$i$位为1,则$i$号点向$log_2^Y+1$连一条边。

代码

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#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+100;
const int maxm=5e5+100;
struct io{
    template<typename T> void read(T &n)
    {
        n=0;
        char c;bool f=0;
        for(;!isdigit(c);c=getchar())if(c=='-')f=1;
        for(;isdigit(c);c=getchar())n=n*10+c-'0';
        if(f)n=-n;
    }
    template<typename T> io operator >> (T &n)
    {
        this->read(n);
        return *this;
    }
}yin;
struct edge{
    int v,w,nxt,del;
}e[maxm<<1];
int tot,head[maxn];
inline void __ADD(int x,int y,int z)
{
    e[++tot].v=y;
    e[tot].w=z;
    e[tot].nxt=head[x];
    head[x]=tot;
}
inline void add(int x,int y,int z)
{
    __ADD(x,y,z);
    __ADD(y,x,z);
}
long long mind1[maxn],mind2[maxn];
bool vis[maxn];
int n,m;
int fa[maxn];
void dij(int t)
{
    memset(vis,0,sizeof(vis));
    for(register int i=2;i<=n;i++)mind1[i]=LLONG_MAX>>2;
    priority_queue<pair<long long,int> >q;
    while(!q.empty())q.pop();
    q.push(make_pair(0,1));
    while(!q.empty())
    {
        register int u=q.top().second;
        q.pop();
        if(vis[u])continue;
        vis[u]=1;
        for(register unsigned int i=head[u];i;i=e[i].nxt)
        {
            if(e[i].del)continue;
            register int v=e[i].v;
            register long long w=(long long)e[i].w+mind1[u];
            if(mind1[v]>w)
            {
                mind1[v]=w;
                if(t)fa[v]=i;
                q.push(make_pair(-mind1[v],v));
            }
        }
    }
}
int in[maxn];
int main()
{
    freopen("rebirth.in","r",stdin);
    freopen("rebirth.out","w",stdout);
    int num;
    yin>>num;
    yin>>n>>m;
    for(int i=1;i<=m;i++)
    {
        int x,y,z;
        yin>>x>>y>>z;
        ++in[x];
        ++in[y];
        if(x^y)add(x,y,z);
    }
    dij(1);
    for(int i=1;i<=n;i++)
    {
        if(in[i]==1){printf("-1 ");continue;}
        e[fa[i]].del=1;
        dij(0);
        if(mind1[i]!=LLONG_MAX>>2)
            printf("%lld ",mind1[i]);
        else printf("-1 ");
        e[fa[i]].del=0;
    }
    return 0;
}