HDU-1814-TwoSAT

題目大意：有(1,2),(3,4), ...(2n-1,2n),n對數，要在每對中選出一個共n個，並且滿足m個約束條件，條件(a,b)表示ab不能同時被選,按字典序輸出答案；

題目解析：如果有a b有約束，那麼建邊A->B^1,B->A^1，字典序最後直接遍歷輸出；

AC代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<stack>
using namespace std;
const int maxn=8010*2;
const int maxm = 20010;
int first[maxn];  
int vv[maxm*4],nxt[maxm*4],S[maxm*4];  
int e,c,n,m;  
bool vis[maxn];  
void addedge(int u,int v)  
{  
    vv[e] = v;  nxt[e] = first[u];  first[u] = e++;  
}  
  
bool dfs(int u) 
{  
    if(vis[u^1])    return 0;  
    if(vis[u])  return 1;  
    vis[u] = 1;  
    S[c++] = u;  
    for(int i = first[u];i != -1;i = nxt[i])  
    {  
        int v = vv[i];  
        if(!dfs(v)) return false;  
    }  
    return true;  
}  
  
bool Judge()  
{  
    for(int i = 0;i < n*2;i+=2)  
    {  
        if(!vis[i] && !vis[i+1])  
        {  
            c = 0;  
            if(!dfs(i))  
            {  
                while(c > 0) vis[S[--c]] = 0;  
                if(!dfs(i+1))   return false;  
            }  
        }  
    }  
    return true;  
}  
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF&&(n+m))
	{
		memset(first,-1,sizeof(first));
		e=0;
		memset(vis,0,sizeof(vis));
		for(int i=0;i<m;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			u--;
			v--; 
			addedge(u,v^1);
			addedge(v,u^1);
		} 
		if(Judge())
		{
			for(int i=0;i<n;i++)
			{
				if(vis[2*i+1]==1)
					printf("%d\n",i*2+2);
				else 
					printf("%d\n",i*2+1);
			}
		}
		else 
			printf("NIE\n");
	}
	return 0;
}