BZOJ2079 [Poi2010]Guilds 【贪心】

题目链接

BZOJ2079

题解

题意就是黑白染色，要求相邻点存在不同颜色的点

显然从一个点出发，相邻点如果没有染色，染不同颜色，那么一个联通块一定会满足要求

证明：在\(dfs\)树上，每个点父亲和它不同色

所以只需检查有无孤立的点

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<map>

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define mp(a,b) make_pair<int,int>(a,b)

#define cls(s) memset(s,0,sizeof(s))

#define cp pair<int,int>

#define LL long long int

using namespace std;

const int maxn = 200005,maxm = 100005,INF = 1000000000;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

int n,m,de[maxn];

int main(){

	n = read(); m = read(); int a,b;

	while (m--){

		a = read(); b = read();

		if (a != b) de[a]++,de[b]++;

	}

	int flag = true;

	REP(i,n) if (!de[i]){flag = false; break;}

	puts(flag ? "TAK" : "NIE");

	return 0;

}