[Trees differential] [dfs] Luogu P4652 One-Way Streets

Title Description

Given an  n- n-spots  m without edges to FIG. M, now we want to this orientation of FIG.

There  P P limitation conditions, each shaped like  (x_i, y_i) ( X I , Y I ), represented by a directed graph in which the new, x_i X I  be able to walk along the edges of some  y_i Y i .

Now you find, whether the direction of each side can be uniquely determined. Please also given side direction can be uniquely determined by these.

 

 

answer

• It is easy to find, if I i have added a rear edge, if there is a ring, which above all B side
• Then we consider how the L and R judgment, it is clear that, after completion of sentence B, the rest are acyclic 
• Says will not because the direction of two different target path 
• This inspired us to use a tree difference, then you know one side of the point 

 

Code

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 const int N=100010;
 7 int n,m,p,cnt,head[N],f2[N],f1[N];
 8 struct edge { int x,y,from,id; }e[N*2];
 9 void init (int x,int y,int z) { e[++cnt].x=x,e[cnt].y=y,e[cnt].id=z,e[cnt].from=head[x],head[x]=cnt; }
10 bool vis[N],vis1[N<<1];
11 char ans[N];
12 void dfs (int x)
13 {
14     vis[x]=true;
15     for (int i=head[x];i!=-1;i=e[i].from)
16     {
17         int y=e[i].y;
18         if (!vis[y]) vis1[e[i].id]=true,dfs(y);
19         else if (!vis1[e[i].id]) ans[e[i].id]='B',vis1[e[i].id]=true,f1[x]++,f1[y]--;
20     }
21 }
22 void dfs1 (int x)
23 {
24     vis[x]=true;
25     for (int i=head[x];i!=-1;i=e[i].from)
26         if (!vis[e[i].y])
27         {
28             int y=e[i].y; dfs1(y);
29             if (f1[y]>0) ans[e[i].id]='B';
30             else if ((f2[y]>0&&(i&1))||(f2[y]<0&&(!(i&1)))) ans[e[i].id]='L';
31             else if ((f2[y]<0&&(i&1))||(f2[y]>0&&(!(i&1)))) ans[e[i].id]='R';
32             f1[x]+=f1[y];f2[x]+=f2[y];
33         }
34 }
35 int main()
36 {
37     scanf("%d%d",&n,&m),memset(head,-1,sizeof(head));
38     for (int i=1,x,y;i<=m;i++) scanf("%d%d",&x,&y),init(x,y,i),init(y,x,i);
39     scanf("%d",&p);
40     for (int i=1,x,y;i<=p;i++) scanf("%d%d",&x,&y),f2[x]++,f2[y]--;
41     memset(vis,false,sizeof(vis)),memset(vis1,false,sizeof(vis1));
42     for (int i=1;i<=n;i++) if (!vis[i]) dfs(i);
43     memset(vis,false,sizeof(vis));
44     for (int i=1;i<=n;i++) if (!vis[i]) dfs1(i);
45     for (int i=1;i<=m;i++) if (!ans[i]) printf("B"); else printf("%c",ans[i]);
46 }