• 洛谷 [P2296] 寻找道路


    反向BFS预处理,求出所有符合题意的点,再正向BFS,(注意对于边权恒为一的点,BFS,比SPFA高效)

    输入时n与m分清###

    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cstdlib>
    #include <queue>
    #include <cstring>
    using namespace std;
    const int MAXN=10005,MAXM=2000005;
    int read(){
    	int rv=0,fh=1;
    	char c=getchar();
    	while(c<'0'||c>'9'){
    		if(c=='-') fh=-1;
    		c=getchar();
    	}
    	while(c>='0'&&c<='9'){
    		rv=(rv<<1)+(rv<<3)+c-'0';
    		c=getchar();
    	}
    	return fh*rv;
    }
    int head[MAXN],rhead[MAXN],n,m,s,t,nume,rnume,tot=-1,path[MAXN];
    bool f[MAXN],use[MAXN];
    struct edge{
    	int to,nxt;
    }e[MAXM],re[MAXM];
    void adde(int from,int to){
    	e[++nume].to=to;
    	e[nume].nxt=head[from];
    	head[from]=nume;
    }
    void radde(int from,int to){
    	re[++rnume].to=to;
    	re[rnume].nxt=rhead[from];
    	rhead[from]=rnume;
    }
    void bfs(){
    	queue <int> q;
    	q.push(t);
    	f[t]=1;
    	while(!q.empty()){
    		int u=q.front();
    		q.pop();
    		for(int i=rhead[u];i;i=re[i].nxt){
    			int v=re[i].to;
    			if(!f[v]){
    				f[v]=1;
    				q.push(v);
    			}
    		}
    	} 
    }
    /*void dfs(int u){
    	f[u]=1;
    	for(int i=rhead[u];i;i=re[i].nxt){
    			int v=re[i].to;
    			if(!f[v]){
    				dfs(v);
    				
    			}
    	}
    }*/
    void bfs2(){
    	queue <int> q;
    	q.push(s);
    	f[s]=1;
    	while(!q.empty()){
    		int u=q.front();
    		q.pop();
    		for(int i=head[u];i;i=e[i].nxt){
    			int v=e[i].to;
    			if(!f[v]&&!use[v]){
    				f[v]=1;
    				q.push(v);
    				path[v]=path[u]+1;
    				if(v==t){
    					tot=path[v];
    					return;
    				}
    			}
    		}
    	}
    }
    int main(){
    //	freopen("in.txt","r",stdin);
    	n=read();m=read();
    	for(int i=1;i<=m;i++){
    		int u=read(),v=read();
    		//cout<<u<<" "<<v<<endl;
    		adde(u,v);radde(v,u);
    	}
    	s=read();t=read();
    	bfs();
    	for(int i=1;i<=n;i++){
    		if(!f[i]){
    			//use[i]=1;
    			for(int j=rhead[i];j;j=re[j].nxt){
    				use[re[j].to]=1;
    			}
    		}
    	}
    	//for(int i=1;i<=n;i++) if(!f[i]) printf("%d ",i);
    	memset(f,0,sizeof(f));
    	//for(int i=1;i<=n;i++) printf("%d ",f[i]);
    	bfs2();
    	cout<<tot;
    //	fclose(stdin);
    	return 0;
    }
    

    P2296

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  • 原文地址:https://www.cnblogs.com/Mr-WolframsMgcBox/p/7868386.html
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