• [POJ 3683] Priest John's Busiest Day


    [题目链接]

               http://poj.org/problem?id=3683

    [算法]

             2-SAT,  用拓扑排序输出可行解

    [代码]

          

    #include <algorithm>  
    #include <bitset>  
    #include <cctype>  
    #include <cerrno>  
    #include <clocale>  
    #include <cmath>  
    #include <complex>  
    #include <cstdio>  
    #include <cstdlib>  
    #include <cstring>  
    #include <ctime>  
    #include <deque>  
    #include <exception>  
    #include <fstream>  
    #include <functional>  
    #include <limits>  
    #include <list>  
    #include <map>  
    #include <iomanip>  
    #include <ios>  
    #include <iosfwd>  
    #include <iostream>  
    #include <istream>  
    #include <ostream>  
    #include <queue>  
    #include <set>  
    #include <sstream>  
    #include <stdexcept>  
    #include <streambuf>  
    #include <string>  
    #include <utility>  
    #include <vector>  
    #include <cwchar>  
    #include <cwctype>  
    #include <stack>  
    #include <limits.h>
    using namespace std;
    #define MAXN 1010
    #define MAXM 1000010
    
    struct edge
    {
            int to,nxt;
    } e[MAXM << 1],ec[MAXM << 1];
    
    int i,j,n,m,SH,SM,TH,TM,timer,top,cnt,tot,ctot;
    int S[MAXN],T[MAXN],D[MAXN],low[MAXN << 1],dfn[MAXN << 1],belong[MAXN << 1],
            stk[MAXN << 1],head[MAXN << 1],chead[MAXN << 1],val[MAXN << 1],opp[MAXN << 1];
    bool instack[MAXN << 1];
    stack<int> s;
    
    inline void addedge(int u,int v)
    {
            tot++;
            e[tot] = (edge){v,head[u]};
            head[u] = tot;
    }
    inline void addcedge(int u,int v)
    {
            ctot++;
            ec[ctot] = (edge){v,chead[u]};
            chead[u] = ctot;
    }
    inline bool illegal(int a,int b,int c,int d)
    {
            if (a >= c && a < d) return true;
            if (b > c && b <= d)  return true;
            if (a <= c && b >= d) return true;
            return false;
    }
    inline void tarjan(int u)
    {
            int i,v;
            low[u] = dfn[u] = ++timer;
            instack[u] = true;
            s.push(u);
            for (i = head[u]; i; i = e[i].nxt)
            {
                    v = e[i].to;
                    if (!dfn[v])
                    {
                            tarjan(v);
                            low[u] = min(low[u],low[v]);
                    } else if (instack[v]) low[u] = min(low[u],dfn[v]);
            }
            if (dfn[u] == low[u])
            {
                    cnt++;
                    do
                    {
                            v = s.top(); 
                            s.pop();
                            belong[v] = cnt;
                            instack[v] = false;
                    } while (v != u);
            }
    }
    inline void topsort()
    {
            int i,u,v;
            queue< int > q;
            static int deg[MAXN << 1];
            for (i = 1; i <= cnt; i++) 
            {
                    deg[i] = 0;
                    val[i] = -1;
            }
            for (u = 1; u <= 2 * n; u++)
            {
                    for (i = head[u]; i; i = e[i].nxt)
                    {
                            v = e[i].to;
                            if (belong[u] != belong[v]) 
                            {
                                    addcedge(belong[v],belong[u]);
                                    deg[belong[u]]++;
                            }
                    }
            }
            for (i = 1; i <= cnt; i++)
            {
                    if (!deg[i])
                            q.push(i);
            }
            while (!q.empty())
            {
                    u = q.front();
                    q.pop();
                    if (val[u] == -1) 
                    {
                            val[u] = 0; 
                            val[opp[u]] = 1;
                    }
                    for (i = chead[u]; i; i = ec[i].nxt)
                    {
                            v = ec[i].to;
                            if (--deg[v] == 0) q.push(v);
                    }
            }
            for (i = 1; i <= n; i++)
            {
                    if (val[belong[i]] == 0) printf("%02d:%02d %02d:%02d
    ",S[i] / 60,S[i] % 60,(S[i] + D[i]) / 60,(S[i] + D[i]) % 60);
                    else printf("%02d:%02d %02d:%02d
    ",(T[i] - D[i]) / 60,(T[i] - D[i]) % 60,T[i] / 60,T[i] % 60);
            }
    }
    
    int main() 
    {
            
            scanf("%d",&n);
            for (i = 1; i <= n; i++)
            {
                    scanf("%d:%d %d:%d %d",&SH,&SM,&TH,&TM,&D[i]);
                    S[i] = SH * 60 + SM;
                    T[i] = TH * 60 + TM;        
            }
            for (i = 1; i <= n; i++)
            {
                    for (j = i + 1; j <= n; j++)
                    {
                            if (illegal(S[i],S[i] + D[i],S[j],S[j] + D[j]))
                            {
                                    addedge(i,j + n);
                                    addedge(j,i + n);
                            }
                            if (illegal(S[i],S[i] + D[i],T[j] - D[j],T[j]))
                            {
                                    addedge(i,j);
                                    addedge(j + n,i + n);
                            }
                            if (illegal(T[i] - D[i],T[i],S[j],S[j] + D[j]))
                            {
                                    addedge(i + n,j + n);
                                    addedge(j,i);
                            }
                            if (illegal(T[i] - D[i],T[i],T[j] - D[j],T[j]))
                            {
                                    addedge(i + n,j);
                                    addedge(j + n,i);
                            }
                    }
            }
            for (i = 1; i <= 2 * n; i++)
            {
                    if (!dfn[i])
                        tarjan(i);
            }
            for (i = 1; i <= n; i++)
            {
                    if (belong[i] == belong[i + n])
                    {
                            printf("NO
    ");
                            return 0;
                    }
                    opp[belong[i]] = belong[i + n];
                    opp[belong[i + n]] = belong[i]; 
            }
            printf("YES
    ");
            topsort();
            
            return 0;
        
    }
  • 相关阅读:
    mysql主从复制的一些东西的整理
    (转载)[我只是认真]聊聊工匠情怀
    Redis运维的一些常用的命令总结
    关于mysql和Apache以及nginx的监控脚本怎么写会比较好的记录
    使用linux的nc来进行文件的传输
    nc检测端口是否正常服务的一个命令
    二维数组去除重复值和array_unique函数
    MySQL的备份的一些策略和方法的总结
    一些容易忘记的小知识点
    关于php多线程的记录
  • 原文地址:https://www.cnblogs.com/evenbao/p/9403250.html
Copyright © 2020-2023  润新知