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来源：互联网发布：golang 场景编辑：程序博客网时间：2024/06/11 18:34

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题意：中文

思路：思路就是当前已经走过的点，则将所有的需要这个点的边全部走一遍，这样的话复杂度就比较低了

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;typedef unsigned long long ull;const int inf=0x3f3f3f3f;const ll INF=0x3f3f3f3f3f3f3f3fll;const int maxn=50010;int head1[maxn],head2[maxn],k1,k2;struct edge{    int x,y,next;};edge E[maxn*2],EE[maxn*2];void add_edge1(int u,int v,int w){    E[k1].x=v;E[k1].y=w;E[k1].next=head1[u];head1[u]=k1++;}void add_edge2(int u,int v,int w){    EE[k2].x=v;EE[k2].y=w;EE[k2].next=head2[u];head2[u]=k2++;}bool vis[maxn];void dfs(int x){    vis[x]=1;    for(int i=head2[x];i!=-1;i=EE[i].next){        if(vis[EE[i].x]&&!vis[EE[i].y]) dfs(EE[i].y);        if(!vis[EE[i].x]&&vis[EE[i].y]) dfs(EE[i].x);    }    for(int i=head1[x];i!=-1;i=E[i].next){        if(vis[E[i].y]&&!vis[E[i].x]) dfs(E[i].x);    }}int main(){    int N,J,M,u,v,c;    scanf("%d%d%d",&N,&J,&M);    memset(head1,-1,sizeof(head1));    memset(head2,-1,sizeof(head2));    k1=k2=0;    for(int i=0;i<M;i++){        scanf("%d%d%d",&u,&v,&c);        add_edge1(u,v,c);add_edge1(v,u,c);        add_edge2(c,u,v);    }    dfs(J);    for(int i=1;i<=N;i++){        printf("%d:",i);        if(vis[i]) printf("Yes\n");        else printf("No\n");    }    return 0;}

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