Bzoj2079 [Poi2010] Guilds

Solution

显然黑点或者白点的适用范围是多于灰点的,所以没有必要使用灰点。然后直接染色即可。

无解的充要条件应该是至少有一个孤立的点。

Code

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#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define maxn 200005
#define put() putchar('\n')
#define Tp template<typename Ty>
#define Ts template<typename Ty,typename... Ar>
using namespace std;
inline void read(int &x){
int f=1;x=0;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
x*=f;
}
namespace Debug{
Tp void _debug(char* f,Ty t){cerr<<f<<'='<<t<<endl;}
Ts void _debug(char* f,Ty x,Ar... y){while(*f!=',') cerr<<*f++;cerr<<'='<<x<<",";_debug(f+1,y...);}
Tp ostream& operator<<(ostream& os,vector<Ty>& V){os<<"[";for(auto& vv:V) os<<vv<<",";os<<"]";return os;}
#define gdb(...) _debug((char*)#__VA_ARGS__,__VA_ARGS__)
}using namespace Debug;
int n,tot;
int Ans[maxn];
vector<int>to[maxn];
inline void dfs(int x,int c) {
Ans[x]=c;++tot;
for (auto y:to[x]) if (!Ans[y]) dfs(y,3-c);
}
signed main(void){
// freopen("1.in","r",stdin);
int i,x,y,m;
read(n);read(m);
for (i=1;i<=m;i++) read(x),read(y),to[x].push_back(y),to[y].push_back(x);
for (i=1;i<=n;i++) if (!Ans[i]) {
int now=tot;
dfs(i,1);
if (now+1==tot) return puts("NIE"),0;
}
puts("TAK");
for (i=1;i<=n;i++) puts(Ans[i]==1?"K":"S");
return 0;
}