Case of Computer Network solution to a problem
The meaning of problems:
Given a point n, m undirected graph edges. There q article which reaches through the route from ti to si. Now you want to give a direction to the free edge assign each graph. Q. Is there a distribution route so that all the answers are to be met.
\ (1≤n, m, q≤2 × 10 ^ 5 \)
Early this question seems no ideas,
but we think about it, if you can find one point to another point there are two completely different paths, then
we can make a forward, a reverse,
no matter how inquiries are anyway.
Wow!
"One point to another there are two completely different paths"?
This is not a double-edge connected components do?
All on the same side of the double inquiry would not have considered.
But the path between the double edge of it?
How to do?
There is no bilateral same what role,
we simply shrink point,
then it became a forest.
I only ask two situations:
1. not in the same tree, the direct No;
2. In the same tree: a look at each side forward is not only required, but also requires the reverse.
The trees do some difference on the line.
Two arrays,
a forward, a reverse.
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+6;
int n,m,q;
int t1,t2,cnt=1,f[N],v[N],dfn[N],low[N],head[N],num2[N];
int deep=0,d[N],top[N],son[N],siz[N],book[N],num[N];
int f2[N],id[N],fa[N],head2[N],tot=0,cnt2=0,LCA;
map<int,int> ha[N];
struct edge{int nxt,to;}e[N<<1],g[N<<1];
inline int read(){
int T=0,F=1; char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-') F=-1; ch=getchar();}
while(ch>='0'&&ch<='9') T=(T<<3)+(T<<1)+(ch-48),ch=getchar();
return F*T;
}
inline void add(int u,int v){e[++cnt].nxt=head[u],e[cnt].to=v,head[u]=cnt;}
inline void add2(int u,int v){g[++cnt2].nxt=head2[u],g[cnt2].to=v,head2[u]=cnt2;}
int getf(int x){return fa[x]==x?x:fa[x]=getf(fa[x]);}
void merge(int x,int y){
x=getf(x),y=getf(y);
if(x==y) return;
fa[x]=y;
}
void tarjan(int u,int fr){
dfn[u]=low[u]=++deep;
for(int i=head[u];i;i=e[i].nxt){
if(i==fr) continue;
if(!dfn[e[i].to]) f[e[i].to]=u,tarjan(e[i].to,i^1),low[u]=min(low[u],low[e[i].to]);
else low[u]=min(low[u],dfn[e[i].to]);
}
}
void dfs(int x,int ff,int o){
f2[x]=ff,d[x]=d[ff]+1,siz[x]=1,id[x]=o; int maxt=0;
for(int i=head2[x];i;i=g[i].nxt) if(g[i].to!=ff){dfs(g[i].to,x,o),siz[x]+=siz[e[i].to]; if(siz[maxt]<siz[g[i].to]) maxt=g[i].to;}
son[x]=maxt;
}
void dfs2(int x,int topx){
top[x]=topx;
if(son[x]) dfs2(son[x],topx);
for(int i=head2[x];i;i=g[i].nxt) if(g[i].to!=f2[x]&&g[i].to!=son[x]) dfs2(g[i].to,g[i].to);
}
void dfs3(int x){
v[x]=1;
for(int i=head2[x];i;i=g[i].nxt) if(g[i].to!=f2[x]) dfs3(g[i].to),num[x]+=num[g[i].to],num2[x]+=num2[g[i].to];
if(num[x]&&num2[x]){printf("No\n"); exit(0);}
}
int lca(int x,int y){
while(top[x]!=top[y]){
if(d[top[x]]<d[top[y]]) swap(x,y);
x=f2[top[x]];
}
if(d[x]>d[y]) swap(x,y);
return x;
}
int main(){
n=read(),m=read(),q=read();
for(int i=1;i<=m;++i) t1=read(),t2=read(),add(t1,t2),add(t2,t1);
for(int i=1;i<=n;++i){if(!dfn[i]) tarjan(i,0); fa[i]=i;}
for(int i=1;i<=n;++i) if(f[i]&&low[i]>dfn[f[i]]) v[i]=1,ha[i][f[i]]=ha[f[i]][i]=1;
for(int i=1;i<=n;++i) for(int j=head[i];j;j=e[j].nxt) if(ha[i].find(e[j].to)==ha[i].end()) merge(i,e[j].to);
for(int i=1;i<=n;++i){
t1=getf(i);
if(!v[i]) continue;
if(!book[fa[i]]) ++tot,book[fa[i]]=tot;
if(!book[fa[f[i]]]) ++tot,book[fa[f[i]]]=tot;
add2(book[fa[i]],book[fa[f[i]]]),add2(book[fa[f[i]]],book[fa[i]]);
}
for(int i=1;i<=tot;++i) if(!id[i]) dfs(i,0,i),dfs2(i,i);
for(int i=1;i<=q;++i){
t1=read(),t2=read(),t1=book[fa[t1]],t2=book[fa[t2]];
if(id[t1]!=id[t2]){printf("No\n"); return 0;}
if(t1==t2) continue;
LCA=lca(t1,t2),++num[t1],--num[LCA],++num2[t2],--num2[LCA];
}
memset(v,0,sizeof(v));
for(int i=1;i<=tot;++i) if(!v[i]) dfs3(i);
printf("Yes\n");
return 0;
}