zoukankan      html  css  js  c++  java
  • poj 3683 Priest John's Busiest Day

    Priest John's Busiest Day

     POJ - 3683 

    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 Sito time Ti. 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 Di 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 Sito Si + Di, or from Ti - Di to Ti). 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 SiTi and DiSi and Ti 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
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #define maxn 2010
    using namespace std;
    int a[maxn],b[maxn],belong[maxn],op[maxn],dfn[maxn],low[maxn],st[maxn];
    bool vis[maxn];
    int head[maxn],head2[maxn],d[maxn],col[maxn],num,num2,ind,cnt,top,scc,n;
    struct node{int to,pre;}e[maxn*maxn],e2[maxn*maxn];
    void Insert(int from,int to){
        e[++num].to=to;
        e[num].pre=head[from];
        head[from]=num;
    }
    void Insert2(int from,int to){
        d[to]++;
        e2[++num2].to=to;
        e2[num2].pre=head2[from];
        head2[from]=num2;
    }
    bool judge(int x,int y){
        if(b[x]<=a[y]||a[x]>=b[y])return 0;
        return 1;
    }
    void build(){
        for(int i=1;i<=n;i++)
            for(int j=i+1;j<=n;j++){
                if(judge(2*i,2*j)){Insert(2*i,2*j-1);Insert(2*j,2*i-1);}
                if(judge(2*i,2*j-1)){Insert(2*i,2*j);Insert(2*j-1,2*i-1);}
                if(judge(2*i-1,2*j)){Insert(2*i-1,2*j-1);Insert(2*j,2*i);}
                if(judge(2*i-1,2*j-1)){Insert(2*i-1,2*j);Insert(2*j-1,2*i);}
            }
    }
    void Tarjan(int u){
        dfn[u]=low[u]=++ind;
        st[++top]=u;vis[u]=1;
        for(int i=head[u];i;i=e[i].pre){
            int v=e[i].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]){
            int now=0;scc++;
            while(now!=u){
                now=st[top--];
                vis[now]=0;
                belong[now]=scc;
            }
        }
    }
    void rebuild(){
        cnt=0;
        for(int x=1;x<=2*n;x++)
            for(int i=head[x];i;i=e[i].pre){
                if(belong[x]!=belong[e[i].to])
                Insert2(belong[e[i].to],belong[x]);
            }
    }
    void dfs(int x){
        if(col[x])return;
        col[x]=-1;
        for(int i=head2[x];i;i=e2[i].pre)
            dfs(e2[i].to);
    }
    void topsort(){
        for(int i=1;i<=scc;i++)
            if(!d[i])st[++top]=i;
        while(top){
            int now=st[top--];
            if(col[now])continue;
            col[now]=1;
            dfs(op[now]);
            for(int i=head2[now];i;i=e2[i].pre){
                d[e2[i].to]--;
                if(!d[e2[i].to])st[++top]=e2[i].to;
            }
        }
    }
    void print(int x){
        printf("%.2d:",x/60);
        printf("%.2d ",x%60);
    }
    int main(){
        scanf("%d",&n);
        int x,y;
        for(int i=1;i<=n;i++){
            scanf("%d:%d",&x,&y);a[2*i]=x*60+y;
            scanf("%d:%d",&x,&y);b[2*i-1]=x*60+y;
            scanf("%d",&x);
            b[2*i]=a[i*2]+x;
            a[2*i-1]=b[i*2-1]-x;
        }
        build();
        for(int i=1;i<=2*n;i++)
            if(!dfn[i])Tarjan(i);
        for(int i=1;i<=n;i++)
            if(belong[2*i]==belong[2*i-1]){
                puts("NO");
                return 0;
            }
        puts("YES");
        rebuild();
        for(int i=1;i<=n;i++){
            op[belong[2*i]]=belong[2*i-1];
            op[belong[2*i-1]]=belong[2*i];
        }
        topsort();
        for(int i=1;i<=n;i++)
            if(col[belong[2*i]]==1)
                print(a[2*i]),print(b[2*i]),puts("");
            else print(a[2*i-1]),print(b[2*i-1]),puts("");
        return 0;
    }
  • 相关阅读:
    从员工到总监,我的7个经验心得(转)
    对上司不满,说还是不说
    老子智慧-大道甚夷
    实际操作中命令 su 与 sudo 的区别(转)
    杨氏矩阵查找元素位置Java实现
    Java实现 蓝桥杯 算法提高 周期字串
    Java实现 蓝桥杯 算法提高 周期字串
    Java实现 蓝桥杯 算法提高 学霸的迷宫
    Java实现 蓝桥杯 算法训练 最大最小公倍数
    Java实现 蓝桥杯 算法训练 最大最小公倍数
  • 原文地址:https://www.cnblogs.com/thmyl/p/8982439.html
Copyright © 2011-2022 走看看