poj3683 Priest John's Busiest Day

输出方案的2-sat

直接比较两个点强联通分量的编号，缩完点的图应该是有向无环图，根据原始做法是反图topsort出解，编号小的说明顺序在后，选择这个点符合定义。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;

struct node
{
    int x,y,next;
}a[8100000];int len,last[2100];
void ins(int x,int y)
{
    len++;
    a[len].x=x;a[len].y=y;
    a[len].next=last[x];last[x]=len;
}
int z,dfn[2100],low[2100];
int top,sta[2100];bool v[2100];
int cnt,bel[2100];
void SCC(int x)
{
    dfn[x]=low[x]=++z;
    sta[++top]=x;v[x]=true;
    for(int k=last[x];k;k=a[k].next)
    {
        int y=a[k].y;
        if(dfn[y]==0)
        {
            SCC(y);
            low[x]=min(low[x],low[y]);
        }
        else if(v[y]==true)
            low[x]=min(low[x],dfn[y]);
    }
    if(low[x]==dfn[x])
    {
        int k;cnt++;
        do
        {
            k=sta[top];top--;
            v[k]=false;
            bel[k]=cnt;
        }while(k!=x);
    }
}

int n;char ss[10];
struct point{int st,ed;}p[2100];
int sc()
{
    scanf("%s",ss+1);int slen=strlen(ss+1);
    int k1=0,k2=0,i=1;
    while(ss[i]!=':') k1=k1*10+ss[i]-'0',i++;
    i++;while(i<=slen)k2=k2*10+ss[i]-'0',i++;
    return k1*60+k2;
}
void pr(int g)
{
    if(g/60<10)printf("0");
    printf("%d",g/60);
    
    printf(":");
    
    g%=60;
    if(g<10)printf("0");
    printf("%d",g);
}
bool together(point p1,point p2)
{
    if(p1.st>p2.st)swap(p1,p2);
    return p1.ed<=p2.st;
}
void conposition()
{
    bool b1,b2;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++) if(i!=j)
        {
            b1=together(p[i],p[j]),
            b2=together(p[i],p[j+n]);
            if(!b1&&!b2)ins(i,i+n);
            else if(!b1)ins(i,j+n);
            else if(!b2)ins(i,j);
            
            b1=together(p[i+n],p[j]),
            b2=together(p[i+n],p[j+n]);
            if(!b1&&!b2)ins(i+n,i);
            else if(!b1)ins(i+n,j+n);
            else if(!b2)ins(i+n,j);
        }
}

int w[1100];
int main()
{
    scanf("%d",&n); int st,ed,L;
    for(int i=1;i<=n;i++)
    {
        st=sc(),ed=sc(),scanf("%d",&L);
        p[i].st=st,p[i].ed=st+L;
        p[i+n].st=ed-L,p[i+n].ed=ed;
    }
    conposition();
    
    z=top=cnt=0;
    for(int i=1;i<=2*n;i++)
        if(dfn[i]==0)SCC(i);
        
    for(int i=1;i<=n;i++)
    {
        if(bel[i]==bel[i+n]){printf("NO\n");return 0;}
        w[i]=bel[i]>bel[i+n];
    }
    printf("YES\n");
    for(int i=1;i<=n;i++)
    {
        if(w[i]==0)
            pr(p[i].st),printf(" "),pr(p[i].ed);
        else
            pr(p[i+n].st),printf(" "),pr(p[i+n].ed);
        printf("\n");
    }
    return 0;
}