POJ-3207 Ikki's Story IV - Panda's Trick 2sat

　　题目链接：http://poj.org/problem?id=3207

　　题意：在一个圆圈上有n个点，现在用线把点两两连接起来，线只能在圈外或者圈内，现给出m个限制，第 i 个点和第 j 个点必须链接在一起，问是否存在可行解。

　　容易想到圈内和圈外分别表示2sat的两种状态，对每一个限制 i 和 j ，考虑所有其它横跨他们的限制，然后连边就可以了。

  1 //STATUS:C++_AC_47MS_6300KB

  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=1010;

 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 nod[N/2][2];

 58 int first[N*2],next[N*N],vis[N*N],S[N*2];

 59 int n,m,mt,cnt;

 60 

 61 struct Edge{

 62     int u,v;

 63 }e[N*N];

 64 

 65 void adde(int a,int b)

 66 {

 67     e[mt].u=a,e[mt].v=b;

 68     next[mt]=first[a];first[a]=mt++;

 69 }

 70 

 71 int dfs(int u)

 72 {

 73     if(vis[u^1])return 0;

 74     if(vis[u])return 1;

 75     int i;

 76     vis[u]=1;

 77     S[cnt++]=u;

 78     for(i=first[u];i!=-1;i=next[i]){

 79         if(!dfs(e[i].v))return 0;

 80     }

 81     return 1;

 82 }

 83 

 84 int Twosat()

 85 {

 86     int i,j;

 87     for(i=0;i<n;i+=2){

 88         if(vis[i] || vis[i^1])continue;

 89         cnt=0;

 90         if(!dfs(i)){

 91             while(cnt)vis[S[--cnt]]=0;

 92             if(!dfs(i^1))return 0;

 93         }

 94     }

 95     return 1;

 96 }

 97 

 98 int main()

 99 {

100  //   freopen("in.txt","r",stdin);

101     int i,j,x,y;

102     while(~scanf("%d%d",&n,&m))

103     {

104         n<<=1;

105         mem(first,-1);mt=0;

106         mem(vis,0);

107         for(i=0;i<m;i++){

108             scanf("%d%d",&nod[i][0],&nod[i][1]);

109             if(nod[i][0]>nod[i][1])swap(nod[i][0],nod[i][1]);

110         }

111 

112         for(i=0;i<m;i++){

113             for(j=i+1;j<m;j++){

114                 if( (nod[j][0]<nod[i][0] && nod[j][1]>nod[i][0] && nod[j][1]<nod[i][1])

115                    || (nod[j][0]>nod[i][0] && nod[j][0]<nod[i][1] && nod[j][1]>nod[i][1])){

116                     x=i<<1,y=j<<1;

117                     adde(x,y^1);

118                     adde(x^1,y);

119                     adde(y,x^1);

120                     adde(y^1,x);

121                 }

122             }

123         }

124 

125         printf("%s\n",Twosat()?"panda is telling the truth...":"the evil panda is lying again");

126 

127     }

128     return 0;

129 }