POJ-3207 Ikki's Story IV - Panda's Trick 2sat
2013-07-07 12:02
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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。
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。
//STATUS:C++_AC_47MS_6300KB #include <functional> #include <algorithm> #include <iostream> //#include <ext/rope> #include <fstream> #include <sstream> #include <iomanip> #include <numeric> #include <cstring> #include <cassert> #include <cstdio> #include <string> #include <vector> #include <bitset> #include <queue> #include <stack> #include <cmath> #include <ctime> #include <list> #include <set> #include <map> using namespace std; //using namespace __gnu_cxx; //define #define pii pair<int,int> #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define PI acos(-1.0) //typedef typedef long long LL; typedef unsigned long long ULL; //const const int N=1010; const int INF=0x3f3f3f3f; const int MOD=5000,STA=100010; const LL LNF=1LL<<60; const double EPS=1e-8; const double OO=1e15; const int dx[4]={-1,0,1,0}; const int dy[4]={0,1,0,-1}; const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; //Daily Use ... inline int sign(double x){return (x>EPS)-(x<-EPS);} template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} template<class T> inline T lcm(T a,T b,T d){return a/d*b;} template<class T> inline T Min(T a,T b){return a<b?a:b;} template<class T> inline T Max(T a,T b){return a>b?a:b;} template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} //End int nod[N/2][2]; int first[N*2],next[N*N],vis[N*N],S[N*2]; int n,m,mt,cnt; struct Edge{ int u,v; }e[N*N]; void adde(int a,int b) { e[mt].u=a,e[mt].v=b; next[mt]=first[a];first[a]=mt++; } int dfs(int u) { if(vis[u^1])return 0; if(vis[u])return 1; int i; vis[u]=1; S[cnt++]=u; for(i=first[u];i!=-1;i=next[i]){ if(!dfs(e[i].v))return 0; } return 1; } int Twosat() { int i,j; for(i=0;i<n;i+=2){ if(vis[i] || vis[i^1])continue; cnt=0; if(!dfs(i)){ while(cnt)vis[S[--cnt]]=0; if(!dfs(i^1))return 0; } } return 1; } int main() { // freopen("in.txt","r",stdin); int i,j,x,y; while(~scanf("%d%d",&n,&m)) { n<<=1; mem(first,-1);mt=0; mem(vis,0); for(i=0;i<m;i++){ scanf("%d%d",&nod[i][0],&nod[i][1]); if(nod[i][0]>nod[i][1])swap(nod[i][0],nod[i][1]); } for(i=0;i<m;i++){ for(j=i+1;j<m;j++){ if( (nod[j][0]<nod[i][0] && nod[j][1]>nod[i][0] && nod[j][1]<nod[i][1]) || (nod[j][0]>nod[i][0] && nod[j][0]<nod[i][1] && nod[j][1]>nod[i][1])){ x=i<<1,y=j<<1; adde(x,y^1); adde(x^1,y); adde(y,x^1); adde(y^1,x); } } } printf("%s\n",Twosat()?"panda is telling the truth...":"the evil panda is lying again"); } return 0; }
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