[ABC245G] Foreign Friends
Problem Statement
There are $N$ people and $K$ nations, labeled as Person $1$, Person $2$, $\ldots$, Person $N$ and Nation $1$, Nation $2$, $\ldots$, Nation $K$, respectively.
Each person belongs to exactly one nation: Person $i$ belongs to Nation $A_i$.
Additionally, there are $L$ popular people among them: Person $B_1$, Person $B_2$, $\ldots$, Person $B_L$ are popular.
Initially, no two of the $N$ people are friends.
For $M$ pairs of people, Takahashi, a God, can make them friends by paying a certain cost: for each $1\leq i\leq M$, he can pay the cost of $C_i$ to make Person $U_i$ and Person $V_i$ friends.
Now, for each $1\leq i\leq N$, solve the following problem.
Can Takahashi make Person $i$ an indirect friend of a popular person belonging to a nation different from that of Person $i$? If he can do so, find the minimum total cost needed. Here, Person $s$ is said to be an indirect friend of Person $t$ when there exists a non-negative integer $n$ and a sequence of people $(u_0, u_1, \ldots, u_n)$ such that $u_0=s$, $u_n=t$, and Person $u_i$ and Person $u_{i+1}$ are friends for each $0\leq i < n$.
Constraints
- $2 \leq N \leq 10^5$
- $1 \leq M \leq 10^5$
- $1 \leq K \leq 10^5$
- $1 \leq L \leq N$
- $1 \leq A_i \leq K$
- $1 \leq B_1<B_2<\cdots<B_L\leq N$
- $1\leq C_i\leq 10^9$
- $1\leq U_i<V_i\leq N$
- $(U_i, V_i)\neq (U_j,V_j)$ if $i \neq j$.
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
$N$ $M$ $K$ $L$ $A_1$ $A_2$ $\cdots$ $A_N$ $B_1$ $B_2$ $\cdots$ $B_L$ $U_1$ $V_1$ $C_1$ $U_2$ $V_2$ $C_2$ $\vdots$ $U_M$ $V_M$ $C_M$
Output
Let $X_i$ defined as follows: $X_i$ is $-1$ if it is impossible to make Person $i$ an indirect friend of a popular person belonging to a nation different from that of Person $i$; otherwise, $X_i$ is the minimum total cost needed to do so. Print $X_1, X_2, \ldots, X_N$ in one line, with spaces in between.
Sample Input 1
4 4 2 2 1 1 2 2 2 3 1 2 15 2 3 30 3 4 40 1 4 10
Sample Output 1
45 30 30 25
Person $1$, $2$, $3$, $4$ belong to Nation $1$, $1$, $2$, $2$, respectively, and there are two popular people: Person $2$ and $3$. Here,
- For Person $1$, the only popular person belonging to a different nation is Person $3$. To make them indirect friends with the minimum cost, we should pay the cost of $15$ to make Person $1$ and $2$ friends and pay $30$ to make Person $2$ and $3$ friends, for a total of $15+30=45$.
- For Person $2$, the only popular person belonging to a different nation is Person $3$. The minimum cost is achieved by making Person $2$ and $3$ friends by paying $30$.
- For Person $3$, the only popular person belonging to a different nation is Person $2$. The minimum cost is achieved by making Person $2$ and $3$ friends by paying $30$.
- For Person $4$, the only popular person belonging to a different nation is Person $2$. To make them indirect friends with the minimum cost, we should pay the cost of $15$ to make Person $1$ and $2$ friends and pay $10$ to make Person $1$ and $4$ friends, for a total of $15+10=25$.
Sample Input 2
3 1 3 1 1 2 3 1 1 2 1000000000
Sample Output 2
-1 1000000000 -1
Note that, for Person $1$, Person $1$ itself is indeed an indirect friend, but it does not belong to a different nation, so there is no popular person belonging to a different nation.
暴力怎么做？枚举每一种国籍 \(i\)，考虑这个国籍的明星带给其他人的影响。建立一个虚拟节点，向每一个这个国籍的明星连一条0边。然后以 \(0\) 为起点跑一次 dijkstra，得到每一个点的最短路，而对于 \(a_i\ne a_j\) 的人，用当前跑出来的最短路更新其答案。复杂度 \(O(mk\log m)\)
为什么我们要枚举国籍？其实就只是为了保证 \(a_i\ne a_j\)，这个作用有点废，考虑二进制优化，什么时候 \(a_i\ne a_j\) 呢？一定存在某一个二进制位不同。枚举每个二进制位，把虚拟节点向所有 \(a_i\) 的这个二进制位为 1 的节点连边，跑完之后，所有这一个二进制位为 0 的节点更新答案。复杂度将为 \(O(m \log^2m)\)
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
struct edge{
int v,nxt,w;
}e[N*3];
int hd[N],a[N],f[N],idx,u,v,w;
long long dis[N],ans[N];
void add_edge(int u,int v,int w)
{
e[++idx]=(edge){v,hd[u],w};
hd[u]=idx;
}
struct node{
int v;
long long w;
bool operator<(const node&n)const{
return w>n.w;
}
};
int n,m,k,l;
priority_queue<node>q;
inline void dijkstra()
{
memset(dis,0x7f,sizeof(dis));
q.push((node){0,0});
dis[0]=0;
while(!q.empty())
{
v=q.top().v;
if(dis[v]!=q.top().w)
{
q.pop();
continue;
}
q.pop();
for(int i=hd[v];i;i=e[i].nxt)
{
if(dis[v]+e[i].w<dis[e[i].v])
{
dis[e[i].v]=dis[v]+e[i].w;
q.push((node){e[i].v,dis[e[i].v]});
}
}
}
}
inline void calc(int i,int t)
{
for(int j=1;j<=n;j++)
if(((a[j]>>i&1)==t)&&f[j])
add_edge(0,j,0);
dijkstra();
hd[0]=0,idx=m<<1;
for(int j=1;j<=n;j++)
if(t!=(a[j]>>i&1))
ans[j]=min(ans[j],dis[j]);
}
int main()
{
memset(ans,0x7f,sizeof(ans));
scanf("%d%d%d%d",&n,&m,&k,&l);
for(int i=1;i<=n;i++)
scanf("%d",a+i);
for(int i=1;i<=l;i++)
scanf("%d",&u),f[u]=1;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
add_edge(u,v,w);
add_edge(v,u,w);
}
for(int i=0;i<17;i++)
{
calc(i,0);
calc(i,1);
}
for(int i=1;i<=n;i++)
printf("%lld ",ans[i]>1e18? -1:ans[i]);
return 0;
}
