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;
}

Poj3683：Priest John's Busiest Day

原文地址：https://www.cnblogs.com/cjoieryl/p/8463664.html