POJ-3207-TwoSAT

題目大意：一個圓上有m對點需要相連，連的方法可以在圓裏外連，問是否可以使得兩兩連線不相交；

題目解析：判斷如果兩個線段如果會有衝突，那麼只能一個在裏面一個在外面；

AC代碼：

#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;
}