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 }
