Acwing 342题解

Acwing 342题解

代码实现

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+5;
typedef long long ll;
#define max(a,b) (a>=b?a:b)
#define min(a,b) (a>=b?b:a)
typedef pair<int,int> pii;
struct node{ //link-table
	int ver,w,next;
}edge[maxn];
int head[maxn],tot;
void add(int u,int v,int w){
	edge[++tot].next=head[u]; edge[tot].ver=v;
	edge[tot].w=w; head[u]=tot;
}
int cnt,pre[maxn];//link-block
void dfs(int x){
	pre[x]=cnt ;
	for(int i=head[x];i;i=edge[i].next){
		int v=edge[i].ver;
		if(!pre[v]){
			dfs(v);
		}
	} return ;
}
int deg[maxn],T,R,P,S;//main-data
int vis[maxn],dist[maxn];//
void solve(int s){
	queue<int> q;
	for(int i=1;i<=cnt;i++){ //point of in-degee=0
		if(deg[i]==0) q.push(i);
	}
	memset(dist,0x3f,sizeof(dist));
	dist[s]=0;//init dist
	while(q.size()){ //topo sort
		int x=q.front(); q.pop();
		priority_queue<pii> pq;
		for(int i=1;i<=T;i++){
			if(pre[i]==x)
				pq.push(make_pair(-dist[i],i) ) ;
		}
		while(pq.size()){
			int nx=pq.top().second; pq.pop();
			if(!vis[nx]){
				vis[nx]=1;
				for(int i=head[nx];i;i=edge[i].next){
					int v=edge[i].ver,w=edge[i].w;
					if(dist[v]>dist[nx]+w){
						dist[v]=dist[nx]+w;
						if(pre[nx]==pre[v])//in a same block
							pq.push(make_pair(-dist[v],v));
					}
					if(pre[nx]!=pre[v]){
						deg[pre[v]]--;
						if(deg[pre[v]]==0)
							q.push(pre[v]) ;
					}
				}
			}
		}
			
	}
}
int main(){
	ios::sync_with_stdio(false); cin.tie(0);cout.tie(0);
	cin>>T>>R>>P>>S;
	while(R--){
		int a,b,c;
		cin>>a>>b>>c;
		add(a,b,c);//road
		add(b,a,c);
	}
	for(int i=1;i<=T;i++){ //divide block
		if(!pre[i]){
			cnt++;
			dfs(i);
		}
	}	
	while(P--){
		int a,b,c; 
		cin>>a>>b>>c; //channel
		add(a,b,c);
		deg[pre[b]]++;
	}
	solve(S);//function
	for(int i=1;i<=T;i++){ //print ans
		if(dist[i]>1e9) cout<<"NO PATH\n";
		else cout<<dist[i]<<endl;
	}
	
	return 0;
}