题目链接：http://codeforces.com/contest/688/problem/C
题意：给一张图，问是否可以划分成二分图。可以的话打印出两个集合。
直接染色，相同颜色放入同一个集合里。如果有单独的点，则任意划分。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rs(a) scanf("%s", a) 44 #define Cin(a) cin >> a 45 #define FRead() freopen("in", "r", stdin) 46 #define FWrite() freopen("out", "w", stdout) 47 #define Rep(i, len) for(int i = 0; i < (len); i++) 48 #define For(i, a, len) for(int i = (a); i < (len); i++) 49 #define Cls(a) memset((a), 0, sizeof(a)) 50 #define Clr(a, x) memset((a), (x), sizeof(a)) 51 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 52 #define lrt rt << 1 53 #define rrt rt << 1 | 1 54 #define pi 3.14159265359 55 #define RT return 56 #define lowbit(x) x & (-x) 57 #define onecnt(x) __builtin_popcount(x) 58 typedef long long LL; 59 typedef long double LD; 60 typedef unsigned long long ULL; 61 typedef pair<int, int> pii; 62 typedef pair<string, int> psi; 63 typedef pair<LL, LL> pll; 64 typedef map<string, int> msi; 65 typedef vector<int> vi; 66 typedef vector<LL> vl; 67 typedef vector<vl> vvl; 68 typedef vector<bool> vb; 69 70 71 typedef struct Edge { 72 int u, v; 73 int next; 74 }Edge; 75 76 const int maxn = 100100; 77 int n, m, ecnt; 78 int color[maxn]; 79 int head[maxn]; 80 Edge edge[maxn*100]; 81 bool flag; 82 vector<int> ret[2]; 83 84 void init() { 85 ecnt = 0; 86 Clr(head, -1); Clr(color, -1); 87 } 88 void adde(int u, int v) { 89 edge[ecnt].u = u; 90 edge[ecnt].v = v; 91 edge[ecnt].next = head[u]; 92 head[u] = ecnt++; 93 } 94 95 void dfs(int u, int c) { 96 if(!flag) return; 97 if(color[u] != -1) { 98 if(color[u] != c) flag = 0; 99 return; 100 } 101 color[u] = c; 102 ret[c].push_back(u); 103 for(int i = head[u]; ~i; i=edge[i].next) { 104 int v = edge[i].v; 105 dfs(v, c^1); 106 } 107 } 108 109 int main() { 110 // FRead(); 111 int u, v; 112 while(~Rint(n) && Rint(m)) { 113 init(); ret[0].clear(); ret[1].clear(); flag = 1; 114 Rep(i, m) { 115 Rint(u); Rint(v); 116 adde(u, v); adde(v, u); 117 } 118 For(i, 1, n+1) { 119 if(color[i] == -1) dfs(i, 0); 120 } 121 if(!flag) { 122 puts("-1"); 123 continue; 124 } 125 Rep(i, 2) { 126 printf("%d ", ret[i].size()); 127 Rep(j, ret[i].size()) printf("%d ", ret[i][j]); 128 printf(" "); 129 } 130 } 131 RT 0; 132 }