Peaceful Commission
Time Limit: 5000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 1814
64-bit integer IO format: %I64d Java class name: Main
The Public Peace Commission should be legislated in Parliament of The Democratic Republic of Byteland according to The Very Important Law. Unfortunately one of the obstacles is the fact that some deputies do not get on with some others.
The Commission has to fulfill the following conditions:
1.Each party has exactly one representative in the Commission,
2.If two deputies do not like each other, they cannot both belong to the Commission.
Each party has exactly two deputies in the Parliament. All of them are numbered from 1 to 2n. Deputies with numbers 2i-1 and 2i belong to the i-th party .
Task
Write a program, which:
1.reads from the text file SPO.IN the number of parties and the pairs of deputies that are not on friendly terms,
2.decides whether it is possible to establish the Commission, and if so, proposes the list of members,
3.writes the result in the text file SPO.OUT.
Input
In the first line of the text file SPO.IN there are two non-negative integers n and m. They denote respectively: the number of parties, 1 <= n <= 8000, and the number of pairs of deputies, who do not like each other, 0 <= m <=2 0000. In each of the following m lines there is written one pair of integers a and b, 1 <= a < b <= 2n, separated by a single space. It means that the deputies a and b do not like each other.
There are multiple test cases. Process to end of file.
Output
The text file SPO.OUT should contain one word NIE (means NO in Polish), if the setting up of the Commission is impossible. In case when setting up of the Commission is possible the file SPO.OUT should contain n integers from the interval from 1 to 2n, written in the ascending order, indicating numbers of deputies who can form the Commission. Each of these numbers should be written in a separate line. If the Commission can be formed in various ways, your program may write mininum number sequence.
Sample Input
3 2 1 3 2 4
Sample Output
1 4 5
Source
解题：2-SAT。只是这题要求输出字典序最小的解。。。只能暴力求解了。
这道题求的是可行解中的最小序列。
我们给结点染色，假设白色是未染色，红色是要选取的结点，蓝色是抛弃的结点。
首先从第一个点开始染色，染成红色，同时将同组另一节点染成蓝色，然后将这个点的所有后继结点也染成红色，同时开一个数组记录都染了哪些结点。如果后来发现某结点的后继是蓝色，说明本次染色失败，因为碰到了矛盾（假如一个结点被选取，那么所有后继肯定也得全部选取）。那么因为刚才染色的时候记录了染色的结点，靠这个数组将结点全部还原回白色。然后从第二个结点开始探索，直到全部染色完毕，最后的红色结点就是答案。
0表示未染色
1表示染成红色
2表示染成蓝色
本题解题思路来自http://blog.163.com/shengrui_step/blog/static/20870918720141201262750/
谢谢Shengrui的精准阐述
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 20010;
18 vector<int>g[maxn];
19 int color[maxn],tmp[maxn],tot,n,m;
20 bool dfs(int u){
21 if(color[u] == 1) return true;
22 if(color[u] == 2) return false;
23 color[u] = 1;
24 tmp[tot++] = u;
25 color[(u-1^1)+1] = 2;
26 for(int i = 0; i < g[u].size(); i++)
27 if(!dfs(g[u][i])) return false;
28 return true;
29 }
30 bool solve(){
31 memset(color,0,sizeof(color));
32 for(int i = 1; i <= n<<1; i++){
33 if(color[i] == 0){
34 tot = 0;
35 if(!dfs(i)){
36 for(int j = 0; j < tot; j++){
37 color[tmp[j]] = 0;
38 color[(tmp[j]-1^1)+1] = 0;
39 }
40 if(!dfs((i-1^1)+1)) return false;
41 }
42 }
43 }
44 return true;
45 }
46 int main() {
47 int x,y;
48 while(~scanf("%d %d",&n,&m)){
49 for(int i = 1; i <= n<<1; i++) g[i].clear();
50 for(int i = 0; i < m; i++){
51 scanf("%d %d",&x,&y);
52 g[x].push_back((y-1^1)+1);
53 g[y].push_back((x-1^1)+1);
54 }
55 if(solve()){
56 for(int i = 1; i <= n<<1; i++)
57 if(color[i] == 1) printf("%d\n",i);
58 }else puts("NIE");
59 }
60 return 0;
61 }