POJ3683——Priest John's Busiest Day-优快云博客

本文介绍了一个基于2-SAT问题的算法，用于解决一个小镇上唯一神父如何合理安排为多对新人主持婚礼祝福仪式的问题。每对新人的婚礼都有固定的时间段，而神父必须在每个婚礼的开始或结束进行特殊仪式。文章提供了一种有效的解决方案，确保神父不会在同一时间出现在两个不同的婚礼上。

Description

John is the only priest in his town. September 1st is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples plan to get married on the blessed day. The i-th couple plan to hold their wedding from time S_i to time T_i. According to the traditions in the town, there must be a special ceremony on which the couple stand before the priest and accept blessings. The i-th couple need D_i minutes to finish this ceremony. Moreover, this ceremony must be either at the beginning or the ending of the wedding (i.e. it must be either from S_i to S_i + D_i, or from T_i - D_i to T_i). Could you tell John how to arrange his schedule so that he can present at every special ceremonies of the weddings.

Note that John can not be present at two weddings simultaneously.

Input

The first line contains a integer N ( 1 ≤ N ≤ 1000).
The next N lines contain the S_i, T_i and D_i. S_i and T_i are in the format of hh:mm.

Output

The first line of output contains "YES" or "NO" indicating whether John can be present at every special ceremony. If it is "YES", output another N lines describing the staring time and finishing time of all the ceremonies.

Sample Input

2
08:00 09:00 30
08:15 09:00 20

Sample Output

YES
08:00 08:30
08:40 09:00

Source

2-sat问题，把一对夫妻的2个时间段看成一个点对，如果与其他人矛盾就连一条边

#include<set>
#include<map>
#include<list>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

const int N=2100;

struct node
{
	int from;
    int to;
    int next;
}edge1[N*N],edge2[N*N];
int head1[N],head2[N];
int tot1,tot2,sccnum,index,Top,cnt;
int DFN[N],low[N];
int stack[N];
int in_deg[N];
int cfl[N];
int toposort[N];
int block[N];
int color[N];
bool instack[N];
int s[1010],t[1010],d[1010];

void addedge1(int from,int to)//原图
{
	edge1[tot1].from=from;
    edge1[tot1].to=to;
    edge1[tot1].next=head1[from];
    head1[from]=tot1++;
}

void addedge2(int from,int to)//缩点后的逆图
{
	edge1[tot2].from=from;
    edge2[tot2].to=to;
    edge2[tot2].next=head2[from];
    head2[from]=tot2++;
}

void tarjan(int u)
{
    DFN[u]=low[u]=++index;
    stack[Top++]=u;
    instack[u]=1;
    for (int i=head1[u];i!=-1;i=edge1[i].next)
    {
        int v=edge1[i].to;
        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])
    {
    	int v;
        sccnum++;
        do
        {
            Top--;
            v=stack[Top];
            block[v]=sccnum;
            instack[v]=0;
        }
        while (v!=u);
    }
}

void topo_sort()
{
    queue<int>qu;
    while (!qu.empty())
        qu.pop();
    for (int i=1;i<=sccnum;i++)//这里已经用到逆图了
        if (in_deg[i]==0)
            qu.push(i);
    while (!qu.empty())
    {
        int u=qu.front();
        if(color[u]==0)
        {
        	color[u]=1;
        	color[cfl[u]]=-1;
        }
        qu.pop();
        for (int i=head2[u];i!=-1;i=edge2[i].next)
        {
            int v=edge2[i].to;
            toposort[++cnt]=v;
            in_deg[v]--;
            if (in_deg[v]==0)
                qu.push(v);
        }
    }
}

void init()
{
	memset(instack,0,sizeof(instack));
	memset(head1,-1,sizeof(head1));
	memset(head2,-1,sizeof(head2));
	memset(DFN,0,sizeof(DFN));
	memset(low,0,sizeof(low));
	memset(color,0,sizeof(color));
	memset(in_deg,0,sizeof(in_deg));
	cnt=0;
	sccnum=0;
	index=0;
	tot1=0;
	tot2=0;
	Top=0;
}

bool judge(int s1,int t1,int s2,int t2)
{
	if(t1<=s2 || t2<=s1)
		return false;
	return true;
}

void out_put(int s,int t)
{
	if(s/60<10)
		printf("0");
	printf("%d:",s/60);
	s%=60;
	if(s<10)
		printf("0");
	printf("%d",s);
	printf(" ");
	if(t/60<10)
		printf("0");
	printf("%d:",t/60);
	t%=60;
	if(t<10)
		printf("0");
	printf("%d\n",t);
}

int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		init();
		int h1,m1,h2,m2;
		bool flag=false;
		for(int i=1;i<=n;i++)
		{
			scanf("%d:%d %d:%d %d",&h1,&m1,&h2,&m2,&d[i]);
			s[i]=h1*60+m1;
			t[i]=h2*60+m2;
		}
		for(int i=1;i<=n;i++)
			for(int j=i+1;j<=n;j++)
			{
				if(judge(s[i],s[i]+d[i],s[j],s[j]+d[j]))
				{
					addedge1(i,j+n);
					addedge1(j,i+n);
				}
				
				if(judge(s[i],s[i]+d[i],t[j]-d[j],t[j]))
				{
					addedge1(i,j);
					addedge1(j+n,i+n);
				}
				if(judge(t[i]-d[i],t[i],s[j],s[j]+d[j]))
				{
					addedge1(i+n,j+n);
					addedge1(j,i);
				}
				if(judge(t[i]-d[i],t[i],t[j]-d[j],t[j]))
				{
					addedge1(i+n,j);
					addedge1(j+n,i);
				}
			}
		for(int i=1;i<=2*n;i++)
			if(DFN[i]==0)
				tarjan(i);
		for(int i=1;i<=n;i++)
		{
			if(block[i]==block[i+n])
			{
				flag=true;
				break;
			}
		}
		if(flag)
		{
			printf("NO\n");
			continue;
		}
		//既然有解，缩点后建立逆图,并且把矛盾块找好 
	   	for(int i=1;i<=sccnum;i++)
		{
			if(block[edge1[i].from]!=block[edge1[i].to])
			{
		 		addedge2(block[edge1[i].from],block[edge1[i].to]);
			}
		}
		for(int i=1;i<=n;i++)
		{
			cfl[block[i]]=block[i+n];
			cfl[block[i+n]]=block[i];
		}
		topo_sort();
		printf("YES\n");
		for(int i=1;i<=n;i++)
		{
			if(color[block[i]]==1)
				out_put(s[i],s[i]+d[i]);
			else
				out_put(t[i]-d[i],t[i]);
		}
	}
	return 0;
}