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;
}
|