Poj3683：Priest John's Busiest Day

题意

n对夫妻要结婚，第i对夫妻结婚的婚礼持续时间为[Si, Ti]，他们会举行一个仪式，仪式时间为Di，这个仪式只能举行在开头或者结尾举行，要么[Si, Si+Di]，要么[Ti-Di, Ti]，然而举行仪式的牧师只有一个，问牧师能否举行完所有仪式

按输入顺序输出方案

手动翻译

Sol

\(2-SAT\)输出一组可行解

这个很烦

\(Tarjan\)缩点成\(DAG\)后再拓扑排序+染色

只传递不选的标记

# include <iostream>

# include <stdio.h>

# include <stdlib.h>

# include <string.h>

# include <math.h>

# include <algorithm>

# include <queue>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(2005);

const int __(2e6 + 5);

IL int Input(){

    RG int x = 0, z = 1; RG char c = getchar();

    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);

    return x * z;

}

int n, first[_], head[_], cnt, num, s[_], t[_], d[_], cho[_];

int S[_], vis[_], dfn[_], low[_], Index, col[_], deg[_], oth[_];

struct Edge{

	int to, next;

} edge[__], dag[__];

queue <int> Q;

IL void Add(RG int u, RG int v){

	edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++;

}

IL void Add_DAG(RG int u, RG int v){

	dag[cnt] = (Edge){v, head[u]}, head[u] = cnt++, ++deg[v];

}

IL void Tarjan(RG int u){

	vis[u] = 1, dfn[u] = low[u] = ++Index, S[++S[0]] = u;

	for(RG int e = first[u]; e != -1; e = edge[e].next){

		RG int v = edge[e].to;

		if(!dfn[v]) Tarjan(v), low[u] = min(low[u], low[v]);

		else if(vis[v]) low[u] = min(low[u], dfn[v]);

	}

	if(dfn[u] != low[u]) return;

	RG int v = S[S[0]--]; col[v] = ++num, vis[v] = 0;

	while(v != u) v = S[S[0]--], col[v] = num, vis[v] = 0;

}

IL void Dfs(RG int u){

	if(cho[u] != -1) return;

	cho[u] = 0;

	for(RG int e = head[u]; e != -1; e = dag[e].next) Dfs(dag[e].to);

}

IL int GetTime(){

	return Input() * 60 + Input();

}

IL void OutTime(RG int x){

	printf("%.2d:%.2d ", x / 60, x % 60);

}

IL int Cross(RG int l1, RG int r1, RG int l2, RG int r2){

	if(l1 > l2) swap(l1, l2), swap(r1, r2);

	return r1 > l2;

}

int main(RG int argc, RG char* argv[]){

	n = Input(), Fill(first, -1), Fill(head, -1), Fill(cho, -1);

	for(RG int i = 1; i <= n; ++i)

		s[i] = GetTime(), t[i] = GetTime(), d[i] = Input();

	for(RG int i = 1; i < n; ++i)

		for(RG int j = i + 1; j <= n; ++j){

			if(Cross(s[i], s[i] + d[i], s[j], s[j] + d[j]))	Add(i, j + n), Add(j, i + n);

			if(Cross(s[i], s[i] + d[i], t[j] - d[j], t[j]))	Add(i, j), Add(j + n, i + n);

			if(Cross(t[i] - d[i], t[i], s[j], s[j] + d[j]))	Add(i + n, j + n), Add(j, i);

			if(Cross(t[i] - d[i], t[i], t[j] - d[j], t[j]))	Add(i + n, j), Add(j + n, i);

		}

	RG int tmp = n << 1; cnt = 0;

	for(RG int i = 1; i <= tmp; ++i) if(!dfn[i]) Tarjan(i);

	for(RG int i = 1; i <= n; ++i){

		if(col[i] == col[i + n]) return puts("NO"), 0;

		oth[col[i]] = col[i + n], oth[col[i + n]] = col[i];

	}

	puts("YES");

	for(RG int i = 1; i <= tmp; ++i)

		for(RG int e = first[i]; e != -1; e = edge[e].next)

			if(col[i] != col[edge[e].to]) Add_DAG(col[edge[e].to], col[i]);

	for(RG int i = 1; i <= num; ++i) if(!deg[i]) Q.push(i);

	while(!Q.empty()){

		RG int u = Q.front(); Q.pop();

		if(cho[u] != -1) continue;

		cho[u] = 1, Dfs(oth[u]);

		for(RG int e = head[u]; e != -1; e = dag[e].next)

			if(!--deg[dag[e].to]) Q.push(dag[e].to);

	}

	for(RG int i = 1; i <= n; ++i){

		if(cho[col[i]]) OutTime(s[i]), OutTime(s[i] + d[i]);

		else OutTime(t[i] - d[i]), OutTime(t[i]);

		puts("");

	}

	return 0;

}