POJ 3683 Priest John's Busiest Day（2-SAT+方案输出）

Priest John's Busiest Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10010		Accepted: 3425		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

题目链接：POJ 3683

挑战程序书上的例题，重新学习了一下2-SAT，一般的2-SAT是给出一些矛盾的条件，然后根据这些矛盾的条件用强连通分量算法来做。

首先要知道把什么东西拆点，一般是去拆具有两个对立面的事物，比如这题就是神父出现在开头和神父出现在结尾，两者就是不能同时发生的对立面，然后记这两个对立面为$x_i$与$\lnot x_i$，显然一般情况下如果a与b矛盾那么就是说“a与b不能同时发生”，转换成标记符号就是$\lnot (a \land b)$，然后把这个式子拆开得到$\lnot a\lor\lnot b$，那么得到这样一个析取范式，我们可以将他转换成蕴含式子，即$(a \to \lnot b)\land(b \to \lnot a)$，这显然可以转换成两条有向边$<a, \lnot b>$与$<b, \lnot a>$，然后对图进行DFS得到强连通分量，然后看$\lnot x_i$与$x_i$是否在同一个强连通分量里，如果在同一个scc中显然是矛盾的，这题的话就第i个时间段和第j个时间段的开头和结束位置共4种关心进行判断，冲突就按上面的方法连边再scc处理即可，如果存在方案，只要判断$x_i$所在的连通分量编号与$\lnot x_i$所在的连通分量编号即可

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <numeric>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 1010;
const int MAXV = N * 2;
const int MAXE = N * N * 4 * 2;
struct edge
{
	int to, nxt;
	edge() {}
	edge(int _to, int _nxt): to(_to), nxt(_nxt) {}
};
edge E[MAXE];
int head[MAXV], tot;
int dfn[MAXV], low[MAXV], belong[MAXV], st[MAXV], top, ts, ins[MAXV], sc;
int s[N], t[N], d[N];

void init()
{
	CLR(head, -1);
	tot = 0;
	CLR(dfn, 0);
	CLR(low, 0);
	top = sc = 0;
	CLR(ins, 0);
	sc = 0;
}
inline void add(int s, int t)
{
	E[tot] = edge(t, head[s]);
	head[s] = tot++;
}
void Tarjan(int u)
{
	low[u] = dfn[u] = ++ts;
	st[top++] = u;
	ins[u] = 1;
	for (int i = head[u]; ~i; i = E[i].nxt)
	{
		int v = E[i].to;
		if (!dfn[v])
		{
			Tarjan(v);
			low[u] = min(low[u], low[v]);
		}
		else if (ins[v])
			low[u] = min(low[u], dfn[v]);
	}
	if (low[u] == dfn[u])
	{
		++sc;
		int v;
		do
		{
			v = st[--top];
			ins[v] = 0;
			belong[v] = sc;
		} while (u != v);
	}
}
int main(void)
{
	int n, i, j;
	while (~scanf("%d", &n))
	{
		init();
		for (i = 1; i <= n; ++i)
		{
			int h1, m1, h2, m2;
			scanf(" %d:%d %d:%d %d", &h1, &m1, &h2, &m2, &d[i]);
			s[i] = h1 * 60 + m1;
			t[i] = h2 * 60 + m2;
		}
		for (i = 1; i <= n; ++i) //n*n*4*2
		{
			for (j = 1; j <= n; ++j)
			{
				if (i == j)
					continue;
				if (min(s[i] + d[i], s[j] + d[j]) > max(s[i], s[j])) //s-s
				{
					add(i, j + n);
					add(j, i + n);
				}
				if (min(s[i] + d[i], t[j]) > max(s[i], t[j] - d[j])) //s-t
				{
					add(i, j);
					add(j + n, i + n);
				}
				if (min(t[i], s[j] + d[j]) > max(t[i] - d[i], s[j])) //t-s
				{
					add(i + n, j + n);
					add(j, i);
				}
				if (min(t[i], t[j]) > max(t[i] - d[i], t[j] - d[j])) //t-t
				{
					add(i + n, j);
					add(j + n, i);
				}
			}
		}
		for (i = 1; i <= (n << 1); ++i)
			if (!dfn[i])
				Tarjan(i);
		int flag = 1;
		for (i = 1; i <= n; ++i)
			if (belong[i] == belong[i + n])
				flag = 0;
		if (!flag)
			puts("NO");
		else
		{
			puts("YES");
			for (i = 1; i <= n; ++i)
			{
				if (belong[i] < belong[i + n])
					printf("%02d:%02d %02d:%02d\n", s[i] / 60, s[i] % 60, (s[i] + d[i]) / 60, (s[i] + d[i]) % 60);
				else
					printf("%02d:%02d %02d:%02d\n", (t[i] - d[i]) / 60, (t[i] - d[i]) % 60, (t[i]) / 60, (t[i]) % 60);
			}
		}
	}
	return 0;
}