[POJ 3683]Priest John's Busiest Day(2-SAT+拓扑排序输出可行解)

题目链接

http://poj.org/problem?id=3683

题目大意

思路

代码

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>

#define MAXE 4100000
#define MAXV 2100

using namespace std;

struct edge
{
    int u,v,next;
}edges1[MAXE],edges2[MAXE];

int head1[MAXV],nCount1=0;
int head2[MAXV],nCount2=0;

int inDegree[MAXV];

void AddEdge1(int U,int V)
{
    edges1[++nCount1].u=U;
    edges1[nCount1].v=V;
    edges1[nCount1].next=head1[U];
    head1[U]=nCount1;
}

void AddEdge2(int U,int V)
{
    inDegree[V]++;
    edges2[++nCount2].u=U;
    edges2[nCount2].v=V;
    edges2[nCount2].next=head2[U];
    head2[U]=nCount2;
}

void add(int a,int typea,int b,int typeb)
{
    AddEdge1((a<<1)+typea,((b<<1)+typeb)^1);
    AddEdge1(((a<<1)+typea)^1,(b<<1)+typeb);
}

int dfn[MAXV],low[MAXV],dfs_time=0;
int stack[MAXV<<1],top=0;
bool inStack[MAXV];
int belong[MAXV],scc_cnt=0;
int opp[MAXV]; //opp[i]=与点i相矛盾的点

void TarjanSCC(int u)
{
    dfn[u]=low[u]=++dfs_time;
    stack[++top]=u;
    inStack[u]=true;
    for(int p=head1[u];p!=-1;p=edges1[p].next)
    {
        int v=edges1[p].v;
        if(!dfn[v])
        {
            TarjanSCC(v);
            low[u]=min(low[u],low[v]);
        }
        else if(inStack[v])
            low[u]=min(low[u],dfn[v]);
    }
    if(dfn[u]==low[u])
    {
        scc_cnt++;
        while(1)
        {
            int now=stack[top];
            inStack[now]=false;
            top--;
            belong[now]=scc_cnt;
            if(now==u) break;
        }
    }
}

int n;
int starttime[MAXV],endtime[MAXV]; //每场婚礼的开始时间与结束时间
int col[MAXV]; //col[i]=点i的颜色
int q[MAXV<<1]; //拓扑排序时用的队列

void TopoSort() //拓扑排序
{
    memset(col,0,sizeof(col));
    int h=0,t=0;
    for(int i=1;i<=scc_cnt;i++) if(!inDegree[i]) q[t++]=i;
    while(h<t)
    {
        int u=q[h++];
        if(!col[u]) //u未着色
        {
            col[u]=1;
            col[opp[u]]=-1;
        }
        for(int p=head2[u];p!=-1;p=edges2[p].next)
        {
            int v=edges2[p].v;
            inDegree[v]--;
            if(!inDegree[v])
                q[t++]=v;
        }
    }
    //下面是选取颜色1的来输出
    for(int i=1;i<=n;i++)
    {
        if(col[belong[i*2]]==-1)
        {
            int sthour=starttime[2*i]/60;
            int stmin=starttime[2*i]%60;
            int edhour=endtime[2*i]/60;
            int edmin=endtime[2*i]%60;
            printf("%02d:%02d %02d:%02d\n",sthour,stmin,edhour,edmin);
        }
        else
        {
            int sthour=starttime[2*i+1]/60;
            int stmin=starttime[2*i+1]%60;
            int edhour=endtime[2*i+1]/60;
            int edmin=endtime[2*i+1]%60;
            printf("%02d:%02d %02d:%02d\n",sthour,stmin,edhour,edmin);
        }
    }
}

bool hasIntersec(int i,int j) //看i和j两个时间段是否相冲突，冲突返回true
{
    if(starttime[i]>=endtime[j]||endtime[i]<=starttime[j]) return false;
    return true;
}

int main()
{
    memset(head1,-1,sizeof(head1));
    memset(head2,-1,sizeof(head2));
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        int hour1,min1,hour2,min2,t;
        scanf("%d:%d %d:%d %d",&hour1,&min1,&hour2,&min2,&t);
        starttime[i*2]=hour1*60+min1;
        endtime[i*2]=hour1*60+min1+t;
        starttime[i*2+1]=hour2*60+min2-t;
        endtime[i*2+1]=hour2*60+min2;
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            if(i==j) continue;
            if(hasIntersec(i*2,j*2)) AddEdge1(i*2,j*2+1);
            if(hasIntersec(i*2,j*2+1)) AddEdge1(i*2,j*2);
            if(hasIntersec(i*2+1,j*2)) AddEdge1(i*2+1,j*2+1);
            if(hasIntersec(i*2+1,j*2+1)) AddEdge1(i*2+1,j*2);
            /*if(hasIntersec(i*2,j*2)) add(i,0,j,0); if(hasIntersec(i*2,j*2+1)) add(i,0,j,1); if(hasIntersec(i*2+1,j*2)) add(i,1,j,0); if(hasIntersec(i*2+1,j*2+1)) add(i,1,j,1);*/
        }
    for(int i=2;i<=2*n+1;i++)
        if(!dfn[i])
            TarjanSCC(i);
    bool flag=true;
    for(int i=1;i<=n;i++)
    {
        if(belong[i*2]==belong[i*2+1])
        {
            flag=false;
            break;
        }
        opp[belong[i*2]]=belong[i*2+1];
        opp[belong[i*2+1]]=belong[i*2];
    }
    if(flag) puts("YES");
    else
    {
        puts("NO");
        return 0;
    }
    for(int i=1;i<=nCount1;i++)
    {
        int u=edges1[i].u,v=edges1[i].v;
        if(belong[u]!=belong[v])
            AddEdge2(belong[u],belong[v]);
    }
    TopoSort();
    return 0;
}