[POJ3683 Priest John's Busiest Day]

[关键字]：2-sat

[题目大意]：有n场婚礼，牧师可以在每场婚礼开始后的d_i分钟主持，也可以在结束前的d_i分钟主持，问是否有一种方案可以让牧师主持所有婚礼。

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[分析]：2-sat求解方案的问题。把每场婚礼看成一组点（A，A'），然后对所有有矛盾的点处理：A->B'(A与B有矛盾)、A->B（A与B’有矛盾）……，然后利用2-sat求解，具体方法可见《由对称性解2-sat问题》

[代码]：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#define MAXN 2100
using namespace std;

struct rec
{
       int st,ed;
}p[MAXN];
int n,tot,sum;
int dfn[MAXN],low[MAXN],ops[MAXN],colour[MAXN],index[MAXN],belong[MAXN];
bool instack[MAXN];
vector<int> e[MAXN],dag[MAXN];

bool fright(int x,int y)
{
     if (p[x].ed<=p[y].st || p[y].ed<=p[x].st) return 0; else return 1;
}

void Init()
{
     scanf("%d",&n);
     int a,b,c,d,t;
     for (int i=0;i<n;i++)
     {
         scanf("%d:%d %d:%d %d",&a,&b,&c,&d,&t);
         int t1=a*60+b;
         int t2=c*60+d;
         p[tot].st=t1,p[tot++].ed=t1+t;
         p[tot].st=t2-t,p[tot++].ed=t2;
     }
     for (int i=0;i<n;i++)
         for (int j=0;j<n;j++)
             if (i!=j)
             {
                      if (fright(2*i,2*j)) e[2*i].push_back(2*j+1);
                      if (fright(2*i,2*j+1)) e[2*i].push_back(2*j);
                      if (fright(2*i+1,2*j)) e[2*i+1].push_back(2*j+1);
                      if (fright(2*i+1,2*j+1)) e[2*i+1].push_back(2*j);
             }
     n=tot;
}

stack<int> s;
void tarjan(int u)
{
     dfn[u]=low[u]=++tot;
     s.push(u);
     instack[u]=1;
     int size=e[u].size();
     for (int i=0;i<size;i++)
     {
         int v=e[u][i];
         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])
     {
                        sum++;
                        int v;
                        do{
                              v=s.top();
                              s.pop();
                              instack[v]=0;
                              belong[v]=sum;
                        }while (v!=u);
     }
}
     
void solve()
{
     sum=tot=0;
     memset(instack,0,sizeof(instack));
     memset(dfn,0,sizeof(dfn));
     memset(low,0,sizeof(low));
     memset(belong,0,sizeof(belong));
     for (int i=0;i<n;i++)
         if (!dfn[i]) tarjan(i);
     for (int i=0;i<n;i+=2)
         if (belong[i]==belong[i+1]) 
            {
                                     printf("NO\n");
                                     return;                                     
            }
         else ops[belong[i]]=belong[i+1],ops[belong[i+1]]=belong[i];
     printf("YES\n");
     memset(index,0,sizeof(index));       
     for (int i=0;i<n;i++)
     {
         int size=e[i].size();
         for (int j=0;j<size;j++)
         {
             int v=e[i][j];
             if (belong[i]!=belong[v]) 
                dag[belong[v]].push_back(belong[i]),index[belong[i]]++;
         }
     }
     queue<int> q;
     for (int i=1;i<=sum;i++)
         if (index[i]==0) q.push(i);
     memset(colour,0,sizeof(colour));
     while (!q.empty())
     {
           int u=q.front();
           q.pop();
           if (colour[u]==0) colour[u]=1,colour[ops[u]]=-1;
           int size=dag[u].size();
           for (int i=0;i<size;i++)
               if (--index[dag[u][i]]==0) q.push(dag[u][i]);
     }
     //for (int i=1;i<=sum;i++) printf("%d %d\n",i,ops[i]);
     int a,b,c,d;
     for (int i=0;i<n;i+=2)
         if (colour[belong[i]]==1)
         {
                          a=p[i].st/60;
                          b=p[i].st%60;
                          c=p[i].ed/60;
                          d=p[i].ed%60;
                          printf("%02d:%02d %02d:%02d\n",a,b,c,d);
         }
         else
         {
             a=p[i+1].st/60;
             b=p[i+1].st%60;
             c=p[i+1].ed/60;
             d=p[i+1].ed%60;
             printf("%02d:%02d %02d:%02d\n",a,b,c,d);
         }
}

int main()
{
    Init();
    solve();
    system("pause");
    return 0;
}