POJ-3207-TwoSAT

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题目大意：一个圆上有m对点需要相连，连的方法可以在圆里外连，问是否可以使得两两连线不相交；

题目解析：判断如果两个线段如果会有冲突，那么只能一个在里面一个在外面；

AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<vector>#include<stack>using namespace std;const int maxn=51010*2;const int maxm = 51010;int first[maxn*2];  int vv[maxm*4],nxt[maxm*4],S[maxm*4];  int e,c,n,m;  bool vis[maxn*2];  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()  {  n=m;    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 x[maxn],y[maxn];bool conflict(int i,int j){bool flag = 0;flag |= ((x[j] > x[i] && x[j] < y[i]) && !(y[j] > x[i] && y[j] < y[i]));flag |= ((y[j] > x[i] && y[j] < y[i]) && !(x[j] > x[i] && x[j] < y[i]));return flag;}int main(){while(scanf("%d%d",&n,&m)!=EOF){memset(first,-1,sizeof(first));e=0;memset(vis,0,sizeof(vis));for(int i=0;i<m;i++)scanf("%d%d",&x[i],&y[i]);for(int i=0;i<m;i++){for(int j=i+1;j<m;j++){if(conflict(i,j)){addedge(i*2,j*2+1);addedge(j*2,i*2+1);addedge(i*2+1,j*2);addedge(j*2+1,i*2);}}}if(Judge())printf("panda is telling the truth...\n");else printf("the evil panda is lying again\n");}return 0;}

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