首页 > 代码库 > 2-SAT poj3207将边看做点
2-SAT poj3207将边看做点
Time Limit: 1000MS | Memory Limit: 131072K | |
Total Submissions: 10238 | Accepted: 3762 |
Description
liympanda, one of Ikki’s friend, likes playing games with Ikki. Today after minesweeping with Ikki and winning so many times, he is tired of such easy games and wants to play another game with Ikki.
liympanda has a magic circle and he puts it on a plane, there are n points on its boundary in circular border: 0, 1, 2, …, n ? 1. Evil panda claims that he is connecting m pairs of points. To connect two points, liympanda either places the link entirely inside the circle or entirely outside the circle. Now liympanda tells Ikki no two links touch inside/outside the circle, except on the boundary. He wants Ikki to figure out whether this is possible…
Despaired at the minesweeping game just played, Ikki is totally at a loss, so he decides to write a program to help him.
Input
The input contains exactly one test case.
In the test case there will be a line consisting of of two integers: n and m (n ≤ 1,000, m ≤ 500). The following m lines each contain two integers ai and bi, which denote the endpoints of the ith wire. Every point will have at most one link.
Output
Output a line, either “panda is telling the truth...
” or “the evil panda is lying again
”.
Sample Input
4 2 0 1 3 2
Sample Output
panda is telling the truth...
将边看成点,然后拆点,一个代表圆内边(以点代边),一个代表园外边.
由约束条件搞出哪些点在所建图中必定要相连,然后跑tajan得到联通块‘
并查集也是可以的
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N=1005; const int M=2000005; const int INF=1000000000; int n,m,tot,head[N]; int dfn[N],low[N],index,instack[N],scc; int top,st[N],fa[N]; int x[N],y[N]; struct edge{ int v,nxt; }e[M]; void init(){ tot=index=scc=top=0; memset(dfn,0,sizeof(dfn)); memset(instack,0,sizeof(instack)); memset(head,-1,sizeof(head)); } void insert(int x,int y){ e[tot].v=y; e[tot].nxt=head[x]; head[x]=tot++; } void Tarjan(int u){ instack[u]=1; st[top++]=u; dfn[u]=low[u]=++index; for(int i=head[u];~i;i=e[i].nxt){ int v=e[i].v; if(!dfn[v]) { Tarjan(v); low[u]=min(low[u],low[v]); } else if(instack[v]) low[u]=min(low[u],dfn[v]); } if(dfn[u]==low[u]){ ++scc;int t; do{ t=st[--top]; instack[t]=0; fa[t]=scc; }while(t!=u); } } void build(){ for(int i=1;i<=m;++i){ scanf("%d%d",&x[i],&y[i]); ++x[i],++y[i]; if(x[i]>y[i]) swap(x[i],y[i]); } for(int i=1;i<=m;++i) for(int j=i+1;j<=m;++j) if((x[i]<=x[j]&&y[i]>=x[j]&&y[i]<=y[j])||(x[i]>=x[j]&&x[i]<=y[j]&&y[i]>=y[j])){ insert(i,j+m); insert(j,i+m); insert(i+m,j); insert(j+m,i); } n=2*m; } bool check(){ for(int i=1;i<=m;++i) if(fa[i]==fa[i+m]) return false; return true; } void solve(){ for(int i=1;i<=n;++i) if(!dfn[i]) Tarjan(i); if(check()) puts("panda is telling the truth..."); else puts("the evil panda is lying again"); } int main(){ scanf("%d%d",&n,&m); init(); build(); solve(); }
2-SAT poj3207将边看做点