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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

題目鏈接：POJ 3683

挑戰程序書上的例題，重新學習了一下2-SAT，一般的2-SAT是給出一些矛盾的條件，然後根據這些矛盾的條件用強連通分量算法來做。

首先要知道把什麽東西拆點，一般是去拆具有兩個對立面的事物，比如這題就是神父出現在開頭和神父出現在結尾，兩者就是不能同時發生的對立面，然後記這兩個對立面為$x_i$與$\lnot x_i$，顯然一般情況下如果a與b矛盾那麽就是說“a與b不能同時發生”，轉換成標記符號就是$\lnot (a \land b)$，然後把這個式子拆開得到$\lnot a\lor\lnot b$，那麽得到這樣一個析取範式，我們可以將他轉換成蘊含式子，即$(\lnot a \to b)\land(\lnot b \to a)$，這顯然可以轉換成兩條有向邊$<\lnot a, b>$與$<\lnot b, a>$，然後對圖進行DFS得到強連通分量，然後看$\lnot x_i$與$x_i$是否在同一個強連通分量裏，如果在同一個scc中顯然是矛盾的，這題的話就第i個時間段和第j個時間段的開頭和結束位置共4種關心進行判斷，沖突就按上面的方法連邊再scc處理即可，如果存在方案，只要判斷$x_i$所在的連通分量編號與$\lnot x_i$所在的連通分量編號即可

代碼：

#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;
}

POJ 3683 Priest John's Busiest Day（2-SAT+方案輸出）