CF 272E Dima and Horses staining, DFS difficulty: 2

Http://codeforces.com/problemset/problem/272/E

Think of hatred as a side,

Because the degree can only be 0,1,2,3, so there are several

0,1 can be put directly

2: There are (in), (0,2) Two cases, the first kind of casually put, the second kind put 0 there

3: There are two cases, (0,3), the first kind puts 1, the second puts 0 there

In other words, there are solutions.

DFS Search can

But at first I felt this was not true, because the time is very complicated, but I never thought of it.

#include <cstdio> #include <vector>using namespace Std;const int Maxn=3e5+5;bool other[maxn];vector<int                > E[maxn];int n,m;void Print () {for (int i=1;i<=n;i++) {if (Other[i]) printf ("1");        else printf ("0"); } puts ("");}        void judge (int s) {int o=0;                for (int i=0;i< (int) e[s].size (); i++) {int t=e[s][i];        if (Other[t]==other[s]) o++;                } if (o>1) {other[s]=!other[s];                        for (int i=0;i< (int) e[s].size (); i++) {int t=e[s][i];                if (Other[t]==other[s]) judge (T);        }}}int Main () {scanf ("%d%d", &n,&m);                for (int i=0;i<m;i++) {int f,t;                scanf ("%d%d", &f,&t);                E[f].push_back (t);        E[t].push_back (f);        } for (int i=1;i<=n;i++) {judge (i);        } print (); return 0;} 

　　

CF 272E Dima and Horses staining, DFS difficulty: 2