SGU 121
题意：给你一张图，问你每个顶点必须有黑白两条边（如果它的边数>=2），问你怎么染色，不行就输出no
收获:你会发现不行的情况只有一个单纯的奇数环的时候，反之我们交替染色即可
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 2e2+5; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a;a=a*a;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,x; bool vis[maxn][maxn]; int ans[maxn][maxn]; int in[maxn]; vector<int> G[maxn]; //不行的情况就是存在一个单纯的奇数环 //奇数环加上一些边的话，要从奇度顶点开始dfs void dfs(int u,int col){ rep(i,0,sz(G[u])){ int v = G[u][i]; if(vis[u][v]) continue; vis[u][v] = vis[v][u] = true; ans[u][v] = ans[v][u] = col; dfs(v,col^1); col^=1; } } bool ok(){ set<int> s; rep(u,1,n+1){ s.clear(); if(in[u] < 2) continue; rep(i,0,sz(G[u])){ int v = G[u][i]; s.insert(ans[u][v]); } if(sz(s) < 2) return false; } return true; } int main(){ scanf("%d",&n); rep(i,1,n+1) { while(scanf("%d",&x)&&x){ G[i].pb(x); in[x]++; } } rep(i,1,n+1) if(in[i] > 1 && (in[i] & 1)) dfs(i,1); rep(i,1,n+1) dfs(i,1); if(!ok()) return puts("No solution"),0; rep(u,1,n+1){ rep(i,0,sz(G[u])){ int v = G[u][i]; printf("%d ",ans[u][v]?1:2); } puts("0"); } return 0; }