POJ 3683.Priest John's Busiest Day 2-SAT

Priest John's Busiest Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10144		Accepted: 3472		Special Judge

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

POJ Founder Monthly Contest – 2008.08.31, Dagger and Facer

 

 

题目链接：http://poj.org/problem?id=3683

题意：有一天n对夫妻同时结婚，但是只有一个神父，需要神父只需要在结婚典礼开始的d分钟或者结束的d分钟，即[s,s+d]或者[t-d,t]。问能不能举办所有的婚礼。

思路：2-SAT

2-SAT：

　　什么是2-SAT呢？就是有一些集合，每个集合中有且仅有两个元素，且不能同时选取两个元素，集合间的元素存在一定的选择关系，求解可行性及可行方案。

算法

1、连边

2、tarjan算法缩点，同一个强联通分量缩成一个点。

3、判可行性，即同一集合中的两个点是否同属一个强连通块

4、缩点建新图，连反边

5、拓扑序，若当前点没有被访问过，则选择该点，不选择其另外的点

（缩点方式改成kosaraju算法（即按照原图拓扑排序的点顺序拓扑排序转置图，最后的强联通分量并且是按照拓扑排序的顺序）可以省略第5步，最后scnno[x]>sccno[¬x]即x为真）

连边

2-SAT算法本身并不难，关键是连边：

a->b即选a必选b。

a、b不能同时选：选了a就要选b'，选了b就要选a'。

a、b必须同时选：选了a就要选b，选了b就要选a，选了a'就要选b'，选了b'就要选a'。

a、b必须选一个：选了a就要选b'，选了b就要选a'，选了a'就要选b，选了b'就要选a。

※a必须选：a'->a。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const int maxn=2e3+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e18+7;
priority_queue<P,vector<P>,greater<P> >q;
struct tt
{
    int s,t,d;
} sign[maxn];
vector<int>G[maxn];
void addedge(int u,int v)
{
    G[u].push_back(v);
}
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt;
stack<int>s;
void dfs(int u)
{
    pre[u]=lowlink[u]=++dfs_clock;
    s.push(u);
    for(int i=0; i<G[u].size(); i++)
    {
        int v=G[u][i];
        if(!pre[v])
        {
            dfs(v);
            lowlink[u]=min(lowlink[u],lowlink[v]);
        }
        else if(!sccno[v])
            lowlink[u]=min(lowlink[u],pre[v]);
    }
    if(lowlink[u]==pre[u])
    {
        scc_cnt++;
        while(true)
        {
            int x=s.top();
            s.pop();
            sccno[x]=scc_cnt;
            if(x==u) break;
        }
    }
}
void find_scc(int n)
{
    dfs_clock=scc_cnt=0;
    memset(sccno,0,sizeof(sccno));
    memset(pre,0,sizeof(pre));
    for(int i=1; i<=n; i++)
        if(!pre[i]) dfs(i);
}
tt ans1[maxn],ans2[maxn];
int cmp(tt x,tt y)
{
    return x.t<y.t;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        int st,sm,tt,tm,d;
        scanf("%d:%d %d:%d %d",&st,&sm,&tt,&tm,&d);
        sign[i].s=st*60+sm,sign[i].t=tt*60+tm,sign[i].d=d;
    }
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            if(i==j) continue;
            if(max(sign[i].s,sign[j].s)<min(sign[i].s+sign[i].d,sign[j].s+sign[j].d))
            {
                addedge(i,j+n);
                addedge(j,i+n);
            }
            if(max(sign[i].s,sign[j].t-sign[j].d)<min(sign[i].s+sign[i].d,sign[j].t))
            {
                addedge(i,j);
                addedge(j+n,i+n);
            }
            if(max(sign[i].t-sign[i].d,sign[j].s)<min(sign[i].t,sign[j].s+sign[j].d))
            {
                addedge(i+n,j+n);
                addedge(j,i);
            }
            if(max(sign[i].t-sign[i].d,sign[j].t-sign[j].d)<min(sign[i].t,sign[j].t))
            {
                addedge(i+n,j);
                addedge(j+n,i);
            }
        }
    }
    find_scc(2*n);
    for(int i=1; i<=n; i++)
    {
        if(sccno[i]==sccno[i+n])
        {
            puts("NO");
            return 0;
        }
    }
    puts("YES");
    for(int i=1; i<=n; i++)
    {
        if(sccno[i]>sccno[i+n])
        {
            ans1[i].s=sign[i].s,ans1[i].t=sign[i].s+sign[i].d;
            ans2[i].s=sign[i].t-sign[i].d,ans2[i].t=sign[i].t;

        }
        else
        {
            ans1[i].s=sign[i].t-sign[i].d,ans1[i].t=sign[i].t;
            ans2[i].s=sign[i].s,ans2[i].t=sign[i].s+sign[i].d;
        }
    }
    sort(ans1+1,ans1+n+1,cmp);
    for(int i=1; i<n; i++)
    {
        if(ans1[i].t>ans1[i+1].s)
        {
            for(int j=1; j<=n; j++)
                printf("%02d:%02d %02d:%02d
",ans2[j].s/60,ans2[j].s%60,ans2[j].t/60,ans2[j].t%60);
            return 0;
        }
    }
    for(int i=1; i<=n; i++)
        printf("%02d:%02d %02d:%02d
",ans1[i].s/60,ans1[i].s%60,ans1[i].t/60,ans1[i].t%60);
    return 0;
}