POJ3683 Priest John's Busiest Day

本题可以看出来是2-sat。

难点在于输出一种可行解。

我们想tarjan在这里的作用是什么？缩点。

何为缩点，即把点都放到一个强连通分量里视为一个块，然后对块跑拓扑即可。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int N=1e4+10;
struct edge{int to,nex;}e[8000006];
struct tim{int a,b;}t[N];
int n,head[N],low[N],dfn[N],opp[N],col[N],vis[N],ans[N],in[N],s[N],len[N],top,num,cnt,ccnt;
vector<int>E[N];
queue<int>q;
bool wrong(int x,int y)
{
    if(t[x].b<=t[y].a||t[y].b<=t[x].a)return 0;
    return 1;
}
void init()
{
    ccnt=num=top=cnt=0;
    memset(head,0,sizeof(head));
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(opp,0,sizeof(opp));
    memset(col,0,sizeof(col));
    memset(vis,0,sizeof(vis));
    memset(in,0,sizeof(in));
    memset(ans,-1,sizeof(ans));
}
void dfs(int x)
{
    s[++top]=x;low[x]=dfn[x]=++cnt;vis[x]=1;
    for(int i=head[x];i;i=e[i].nex)
    {
        int y=e[i].to;
        if(!dfn[y]) 
        {
            dfs(y);low[x]=min(low[x],low[y]);
        }
        else if(vis[y])low[x]=min(low[x],dfn[y]);
    }
    if(low[x]==dfn[x])
    {
        ++num;int a;
        while(s[top+1]!=x)
        {
            int a=s[top--];
            col[a]=num;
            vis[a]=0;
        }
        
    }
    return;
}
void solve()
{
    for(int i=1;i<=n*2;++i)
    {
        if(!dfn[i])dfs(i);
    }
    return;
}
void add(int x,int y)
{
    e[++ccnt].to=y;e[ccnt].nex=head[x];head[x]=ccnt;
}
void toposort()
{
    for(int i=1;i<=2*n;++i)
    {
        for(int j=head[i];j;j=e[j].nex)
        {
            int y=e[j].to;
            if(col[y]==col[i])continue;
            E[col[y]].push_back(col[i]);
            in[col[i]]++;
        }
    }
    for(int i=1;i<=num;++i)
    {
        if(!in[i])q.push(i);
    }
    while(!q.empty())
    {
        int x=q.front();q.pop();
        if(ans[x]==-1)
        {
            ans[x]=1;ans[opp[x]]=0;
        }
        for(int i=E[x].size()-1;i>=0;--i)
        {
            in[E[x][i]]--;
            if(in[E[x][i]]==0)q.push(E[x][i]);
        }
    }
    return;
}
int main()
{
    while(~scanf("%d",&n))
    {
        init();int a,b,c,d;
        for(int i=1;i<=n;++i)
        {
            scanf("%d:%d %d:%d %d",&a,&b,&c,&d,&len[i]);
            t[i].a=a*60+b;
            t[i+n].b=c*60+d;
            t[i].b=t[i].a+len[i];
            t[i+n].a=t[i+n].b-len[i];
        }
        for(int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)
        {
            if(i==j)continue;
            if(wrong(i,j))add(i,j+n);
            if(wrong(i,j+n))add(i,j);
            if(wrong(i+n,j))add(i+n,j+n);
            if(wrong(i+n,j+n))add(i+n,j);
        }
        solve();
        for(int i=1;i<=n;++i)
        {
            if(col[i]==col[i+n]){
                puts("NO");return 0;
            }
            opp[col[i]]=col[i+n];
            opp[col[i+n]]=col[i];
        }
        puts("YES");
        toposort();
        for(int i=1;i<=n;++i)
        {
            if(ans[col[i]]==1)    printf("%.2d:%.2d %.2d:%.2d
",t[i].a/60,t[i].a%60,t[i].b/60,t[i].b%60);
            else     printf("%.2d:%.2d %.2d:%.2d
",t[i+n].a/60,t[i+n].a%60,t[i+n].b/60,t[i+n].b%60);
        }
    }
    return 0;
}