poj 2723 Get Luffy Out

题目链接:http://poj.org/problem?id=2723

题目大意:
有N对钥匙(每一对只能选一把),M个门,每个门可用两把钥匙(这两把可能一样,可能不一样,只要有其中一把便可开门),门按给定的顺序排下来,必须开完1号门才能开2号门....直到开完,问最多能开几个门.

题目思路:
相对明显的2-SAT.
设有一把钥匙x,x表示选,~x表示不选.
如果有一对钥匙a,b,则a->~b,b->~a,即a,b不同时存在.
如果有一个门的所需要的钥匙是a,b,则~a->b,~b->a,即a,b不同时不存在.

代码:

#include <stdlib.h>#include <string.h>#include <stdio.h>#include <ctype.h>#include <math.h>#include <time.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <string>#include <iostream>#include <algorithm>using namespace std;#define ull unsigned __int64//#define ll __int64//#define ull unsigned long long#define ll long long#define son1 New(p.xl,xm,p.yl,ym),(rt<<2)-2#define son2 New(p.xl,xm,min(ym+1,p.yr),p.yr),(rt<<2)-1#define son3 New(min(xm+1,p.xr),p.xr,p.yl,ym),rt<<2#define son4 New(min(xm+1,p.xr),p.xr,min(ym+1,p.yr),p.yr),rt<<2|1#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define middle (l+r)>>1#define MOD 1000000007#define esp (1e-8)const int INF=0x3F3F3F3F;const double DINF=10000.00;//const double pi=acos(-1.0);const int M=5010;int min(int x,int y){return x<y? x:y;}int max(int x,int y){return x>y? x:y;}void swap(int& x,int& y){int t=x;x=y;y=t;}int T,cas;vector<int>init[M];int low[M],pre[M],idx[M],ss[M],Hash[M];int n,nn,Index,top,m,N;void dfs(int u){low[u]=pre[u]=++Index;ss[++top]=u;for(int i=0;i<init[u].size();i++){int v=init[u][i];if(pre[v]==-1) dfs(v),low[u]=min(low[u],low[v]);else if(idx[v]==-1) low[u]=min(low[u],pre[v]);}if(low[u]==pre[u]){int v=-1;++nn;while(u!=v) idx[v=ss[top--]]=nn;}}void Tarjan(){memset(idx,-1,sizeof(int)*(n+1));memset(pre,-1,sizeof(int)*(n+1));Index=top=nn=0,ss[0]=-1;for(int u=1;u<=n;u++) if(pre[u]==-1) dfs(u);}bool _sof(){Tarjan();for(int i=1;i<=n;i+=2) if(idx[i]==idx[i+1]) return false;return true;}void _init(){int u,uu,v,vv,i;for(i=1,n=N<<2;i<=n;i++) init[i].clear();for(i=0;i<N;i++){scanf("%d%d",&u,&v);u=u<<1|1,uu=u+1;v=v<<1|1,vv=v+1;init[u].push_back(vv);init[v].push_back(uu);}int ret=0;for(i=1;i<=m;i++){scanf("%d%d",&u,&v);u=u<<1|1,uu=u+1;v=v<<1|1,vv=v+1;init[uu].push_back(v);init[vv].push_back(u);if(_sof())  ret=i;}printf("%d\n",ret);}int main(){//freopen("1.in","r",stdin);//freopen("1.out","w",stdout);//_init()//scanf("%d",&T);for(cas=1;cas<=T;cas++) _init();while(scanf("%d%d",&N,&m) && (N||m))  _init();return 0;}