[CF688C]NP-Hard Problem（二分图染色）

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\n", ret[i].size());
127             Rep(j, ret[i].size()) printf("%d ", ret[i][j]);
128             printf("\n");
129         }
130     }
131     RT 0;
132 }

 

posted @ 2016-08-24 23:50  Kirai  阅读(218)  评论(0)  编辑  收藏  举报