HDU-1814 Peaceful Commission 2sat

  题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1814

  简单的2sat题。

  1 //STATUS:C++_AC_390MS_996KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //using namespace __gnu_cxx;
 25 //define
 26 #define pii pair<int,int>
 27 #define mem(a,b) memset(a,b,sizeof(a))
 28 #define lson l,mid,rt<<1
 29 #define rson mid+1,r,rt<<1|1
 30 #define PI acos(-1.0)
 31 //typedef
 32 typedef long long LL;
 33 typedef unsigned long long ULL;
 34 //const
 35 const int N=8010;
 36 const int INF=0x3f3f3f3f;
 37 const int MOD=5000,STA=100010;
 38 const LL LNF=1LL<<60;
 39 const double EPS=1e-8;
 40 const double OO=1e15;
 41 const int dx[4]={-1,0,1,0};
 42 const int dy[4]={0,1,0,-1};
 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 44 //Daily Use ...
 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 55 //End
 56 
 57 int first[N*2],next[20000*2],vis[N*2],S[N*2];
 58 int n,m,mt,cnt;
 59 
 60 struct Edge{
 61     int u,v;
 62 }e[20000*2];
 63 
 64 void adde(int a,int b)
 65 {
 66     e[mt].u=a,e[mt].v=b;
 67     next[mt]=first[a];first[a]=mt++;
 68 }
 69 
 70 int dfs(int u)
 71 {
 72     if(vis[u^1])return 0;
 73     if(vis[u])return 1;
 74     int i;
 75     vis[u]=1;
 76     S[cnt++]=u;
 77     for(i=first[u];i!=-1;i=next[i]){
 78         if(!dfs(e[i].v))return 0;
 79     }
 80     return 1;
 81 }
 82 
 83 int Twosat()
 84 {
 85     int i,j;
 86     for(i=0;i<n;i+=2){
 87         if(vis[i] || vis[i^1])continue;
 88         cnt=0;
 89         if(!dfs(i)){
 90             while(cnt)vis[S[--cnt]]=0;
 91             if(!dfs(i^1))return 0;
 92         }
 93     }
 94     return 1;
 95 }
 96 
 97 int main()
 98 {
 99  //   freopen("in.txt","r",stdin);
100     int i,j,a,b;
101     while(~scanf("%d%d",&n,&m))
102     {
103         n<<=1;
104         mem(first,-1);mt=0;
105         mem(vis,0);
106         for(i=0;i<m;i++){
107             scanf("%d%d",&a,&b);
108             a--,b--;
109             adde(a,b^1);
110             adde(b,a^1);
111         }
112 
113         if(Twosat()){
114             for(i=0;i<n;i+=2){
115                 printf("%d\n",(vis[i]?i:i^1)+1);
116             }
117         }
118         else puts("NIE");
119     }
120     return 0;
121 }

 

posted @ 2013-07-07 11:46  zhsl  阅读(416)  评论(1编辑  收藏  举报