poj 3683 2-sat

poj 3683  2-sat 
算法很容易想到
建图：
将每个婚礼可行2个区间建立左右界
N个婚礼 2*N个区间 
然后根据区间相交的矛盾建立新边
这里求解可行解比较复杂

最好自己写一下模板

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#include<vector>#include<sstream>#include<string>#include<climits>#include<stack>#include<set>#include<bitset>#include<cmath>#include<deque>#include<map>#include<queue>#define iinf 2000000000#define linf 1000000000000000000LL#define dinf 1e200#define eps 1e-11#define all(v) (v).begin(),(v).end()#define sz(x)  x.size()#define pb push_back#define mp make_pair#define lng long long#define sqr(a) ((a)*(a))#define pii pair<int,int>#define pll pair<lng,lng>#define pss pair<string,string>#define pdd pair<double,double>#define X first#define Y second#define pi 3.14159265359#define ff(i,xi,n) for(int i=xi;i<=(int)(n);++i)#define ffd(i,xi,n) for(int i=xi;i>=(int)(n);--i)#define ffl(i,r) for(int i=head[r];i!=-1;i=edge[i].next)#define cc(i,j) memset(i,j,sizeof(i))#define two(x)((lng)1<<(x))#define N 2111#define M 1000000#define lson l , mid , rt << 1#define rson mid + 1 , r , rt << 1 | 1#define Mod  n#define Pmod(x) (x%Mod+Mod)%Modusing namespace std;typedef vector<int>  vi;typedef vector<string>  vs;typedef unsigned int uint;typedef unsigned lng ulng;template<class T> inline void checkmax(T &x,T y){if(x<y) x=y;}template<class T> inline void checkmin(T &x,T y){if(x>y) x=y;}template<class T> inline T Min(T x,T y){return (x>y?y:x);}template<class T> inline T Max(T x,T y){return (x<y?y:x);}template<class T> T gcd(T a,T  b){return (a%b)==0?b:gcd(b,a%b);}template<class T> T lcm(T a,T b){return a*b/gcd(a,b);}template<class T> T Abs(T a){return a>0?a:(-a);}template<class T> inline T lowbit(T n){return (n^(n-1))&n;}template<class T> inline int countbit(T n){return (n==0)?0:(1+countbit(n&(n-1)));}template<class T> inline bool isPrimeNumber(T n){if(n<=1)return false;for (T i=2;i*i<=n;i++) if (n%i==0) return false;return true;}template<class T> inline T Minmin(T a,T b,T c,T d){return Min(Min(a,b),Min(c,d));}struct ps{    int l,r;}so[N*2];bool cross(int  p,int  q){    return  (so[p].l<=so[q].l)?so[p].r>so[q].l:so[q].r>so[p].l;}struct tarjan_seg{    int head[N],tot,id[N],pre[N],low[N],stack[N],top,n,Index,nn,newid,newhead[N],*cur,contain[N];    struct pp   {        int v,next;    }edge[N*2000];    void add(int u,int v)    {        edge[tot].v=v;        edge[tot].next=cur[u];        cur[u]=tot++;    }    void init()    {        cc(head,-1);        cc(newhead,-1);        tot=0;        newid=0;        cur=head;        cc(pre,-1);        cc(low,-1);        top=0;        cc(id,-1);        Index=0;    }    void tarjan(int r,int p)    {        bool xx=false;        pre[r]=low[r]=++Index;        stack[++top]=r;        ffl(i,r)        {            int v=edge[i].v;            if(v==p&&xx==0)            {                xx=1;                continue;            }            if(pre[v]==-1)            {                tarjan(v,r);                checkmin(low[r],low[v]);            }            else            if(id[v]==-1)            checkmin(low[r],pre[v]);        }        if(pre[r]==low[r])        {            int v;            ++newid;            do            {                v=stack[top--];                id[v]=newid;            }while(v!=r);        }    }    void solve()    {        ff(i,1,n)        if(pre[i]==-1)        tarjan(i,-1);    }    void build_map()    {        int  hash[N]={};        cur=newhead;        ff(i,1,n)        {            ffl(j,i)            {                int v=edge[j].v;                if(id[i]!=id[v]&&hash[id[i]]!=id[v])                add(id[v],id[i]),hash[id[i]]=id[v];            }        }    }    void solve_contain()    {        cc(contain,-1);        cur=contain;        ff(i,1,n)        add(id[i],i);    }    bool two_sat()    {        for(int i=1;i<=n;i+=2)        if(id[i]==id[i+1])        return false;        return true;    }    void dfs(int r)    {        pre[r]=++Index;        for(int i=newhead[r];i!=-1;i=edge[i].next)         if(pre[edge[i].v]==-1) dfs(edge[i].v);        low[++newid]=r;    }    void color_dfs(int r)    {        pre[r]=2;        for(int i=newhead[r];i!=-1;i=edge[i].next)        if(pre[edge[i].v]==-1) color_dfs(edge[i].v);    }    void Color()    {        cc(pre,-1);       for(int i=nn;i>0;--i)       if(pre[low[i]]==-1)       {          int a=edge[contain[low[i]]].v;          int b;          b=((a+1)/2)*4-1-a;           pre[low[i]]=1;           if(pre[id[b]]==-1)           color_dfs(id[b]);       }    }    void topo_sort()    {        nn=newid;        Index=0;        newid=0;        cc(pre,-1);        cc(low,-1);        ff(i,1,nn)        if(pre[i]==-1)        dfs(i);    }    void taganswer()    {        cc(low,-1);        ff(i,1,nn)       if(pre[i]==1) {         for(int j=contain[i];j!=-1;j=edge[j].next)            low[edge[j].v]=1;        }    }    void print_answer()    {        if(!two_sat())        {printf("NO\n");        return ;        }        printf("YES\n");        solve_contain();        build_map();        topo_sort();       Color();       taganswer();        ff(i,1,n)        {            if(low[i]==1)            {                printf("%02d:%02d %02d:%02d\n",so[i].l/60,so[i].l%60,so[i].r/60,so[i].r%60);            }        }    }};int na;tarjan_seg G;int main(){    while(scanf("%d",&na)==1)    {        G.init();        G.n=2*na;        ff(i,1,na)        {            int h1,h2,m1,m2,d;            scanf("%d:%d %d:%d %d",&h1,&m1,&h2,&m2,&d);            h1=h1*60+m1;            h2=h2*60+m2;            so[i*2-1].l=h1,so[i*2-1].r=h1+d;            so[i*2].l=h2-d,so[i*2].r=h2;        }        ff(i,1,2*na)        ff(j,i+1,2*na)        {            int u=(i+1)/2,v=(j+1)/2;            if(u==v) continue;            else            if(cross(i,j))            {                G.add(i,4*v-1-j);                G.add(j,4*u-1-i);            }        }        G.solve();        G.print_answer();    }    return 0;}