poj3683

题解：

2-sat

然后不用拓扑

直接强连通的逆序就可以了

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=2005;
int flag[N],ans,l,n;
int ne[N*N],fi[N],a[N],b[N],zz[N*N],num,t,zhan[N],dfn[N],low[N],an[N];
void jb(int x,int y)
{
    ne[++num]=fi[x];
    fi[x]=num;
    zz[num]=y;
}
void dfs(int x)
{
    low[x]=dfn[x]=++l;
    zhan[++t]=x;
    flag[x]=true;
    for (int i=fi[x];i!=0;i=ne[i])
     {
         if (an[zz[i]])continue;
        if(!dfn[zz[i]])dfs(zz[i]);
        if(!flag[zz[i]])low[x]=min(low[x],dfn[zz[i]]);else
        low[x]=min(low[x],low[zz[i]]);
     }
    if (dfn[x]==low[x])
     {
         ans++;
         while (zhan[t]!=x)
          {
              flag[zhan[t]]=false;
              an[zhan[t--]]=ans;
          }
         an[zhan[t--]]=ans;
         flag[x]=false;
     }
}
int read()
{
    int x=0;char c;
    for (;c<‘0‘||c>‘9‘;c=getchar());
    for (;c>=‘0‘&&c<=‘9‘;c=getchar())x=x*10+c-48;
    return x;
}
int pd(int x,int y)
{
    if (a[x]>a[y])swap(x,y);
    return b[x]>a[y];
}
void sget(int x)
{
    if (x<10)printf("0");
    printf("%d",x);
}
int main()
{
    scanf("%d",&n);
    for (int i=0;i<n;i++)
     {
         a[i]=read()*60+read();
         b[i+n]=read()*60+read();
         b[i]=read();a[i+n]=b[i+n]-b[i];
         b[i]+=a[i];
     }
    for (int i=0;i<2*n;i++)
     for (int j=i+1;j<2*n;j++)
      if (i%n!=j%n&&pd(i,j))
       jb(i,(j+n)%(2*n)),jb(j,(i+n)%(2*n));
    for (int i=0;i<2*n;i++)
     if (!dfn[i])dfs(i);
    for (int i=0;i<n;i++)
     if (an[i]==an[i+n])
      {
          puts("NO");
          return 0;
      }
    puts("YES");
    for (int i=0;i<n;i++)
     if (an[i]<an[i+n])
      {
          sget(a[i]/60);printf(":");
          sget(a[i]%60);printf(" ");
          sget(b[i]/60);printf(":");
          sget(b[i]%60);
          puts("");
      }
     else
      {
          sget(a[i+n]/60);printf(":");
          sget(a[i+n]%60);printf(" ");
          sget(b[i+n]/60);printf(":");
          sget(b[i+n]%60);
          puts("");
      }
    return 0;
}