POJ1904(有向图缩点+输入输出挂参考)

King's Quest

Time Limit: 15000MS		Memory Limit: 65536K
Total Submissions: 8311		Accepted: 3017
Case Time Limit: 2000MS

Description

Once upon a time there lived a king and he had N sons. And there were N beautiful girls in the kingdom and the king knew about each of his sons which of those girls he did like. The sons of the king were young and light-headed, so it was possible for one son to like several girls. 

So the king asked his wizard to find for each of his sons the girl he liked, so that he could marry her. And the king's wizard did it -- for each son the girl that he could marry was chosen, so that he liked this girl and, of course, each beautiful girl had to marry only one of the king's sons. 

However, the king looked at the list and said: "I like the list you have made, but I am not completely satisfied. For each son I would like to know all the girls that he can marry. Of course, after he marries any of those girls, for each other son you must still be able to choose the girl he likes to marry." 

The problem the king wanted the wizard to solve had become too hard for him. You must save wizard's head by solving this problem. 

Input

The first line of the input contains N -- the number of king's sons (1 <= N <= 2000). Next N lines for each of king's sons contain the list of the girls he likes: first Ki -- the number of those girls, and then Ki different integer numbers, ranging from 1 to N denoting the girls. The sum of all Ki does not exceed 200000. 

The last line of the case contains the original list the wizard had made -- N different integer numbers: for each son the number of the girl he would marry in compliance with this list. It is guaranteed that the list is correct, that is, each son likes the girl he must marry according to this list. 

Output

Output N lines.For each king's son first print Li -- the number of different girls he likes and can marry so that after his marriage it is possible to marry each of the other king's sons. After that print Li different integer numbers denoting those girls, in ascending order.

Sample Input

4
2 1 2
2 1 2
2 2 3
2 3 4
1 2 3 4

Sample Output

2 1 2
2 1 2
1 3
1 4
题意:一个国王有N个王子。一共有N个女孩，每个王子可以喜欢多个女孩，但只能取一个女孩。给定一个参考结婚列表,问每个王子可分别与哪几个女孩结婚。
思路：王子与女孩之间建立有向图,再根据参考结婚列表建立反向边,那么与王子处于同一个连通分量的女孩且是王子喜欢的可以和王子结婚。
附输入输出挂

#include"cstdio"
#include"cstring"
#include"algorithm"
using namespace std;
const int MAXN=4010;
struct Edge{
    int to,next;
}es[210000];
int V;
void Scan(int &val)
{
    char ch;
    int x=0;
    bool flag=true;
    ch=getchar();
    if(ch=='-')    flag=false;
    else if('0'<=ch&&ch<='9')    x=(ch-'0');
    while((ch=getchar())&&'0'<=ch&&ch<='9')
        x=x*10+ch-'0';
    val=(flag==true)?x:-x;
}
void Print(int x)
{    
    if(x>9)    Print(x/10);
    putchar(x%10+'0');
}
int head[MAXN],tot;
void add_edge(int u,int v)
{
    es[tot].to=v;
    es[tot].next=head[u];
    head[u]=tot++;
}
int index;
int dfn[MAXN],low[MAXN];
int stack[MAXN],top;
int cpnt[MAXN],cnt;
bool instack[MAXN];
void tarjan(int u)
{
    instack[u]=true;
    stack[top++]=u;
    dfn[u]=low[u]=++index;
    for(int i=head[u];i!=-1;i=es[i].next)
    {
        int v=es[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])
    {
        int v;
        cnt++;
        do{
            v=stack[--top];
            instack[v]=false;
            cpnt[v]=cnt;
        }while(u!=v);
    }
}
int ans[MAXN];
void solve()
{
    memset(ans,0,sizeof(ans));
    for(int i=1;i<=V+V;i++)    
        if(!dfn[i])    tarjan(i);
            
    for(int i=1;i<=V;i++)
    {
        int counter=0;
        for(int j=head[i];j!=-1;j=es[j].next)
        {
            int v=es[j].to;
            if(cpnt[v]==cpnt[i])    ans[counter++]=v-V;
        }
        sort(ans,ans+counter);
        Print(counter);
        for(int j=0;j<counter;j++)    putchar(' '),Print(ans[j]);
        putchar('
');
    }
}

int main()
{
    while(scanf("%d",&V)!=EOF)
    {
        tot=index=top=cnt=0;
        memset(head,-1,sizeof(head));
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(instack,false,sizeof(instack));
        memset(cpnt,0,sizeof(cpnt));
        for(int i=1;i<=V;i++)
        {
            int k;
            Scan(k);
            while(k--)
            {
                int v;
                Scan(v);
                add_edge(i,V+v);
            }
        }    
        for(int i=1;i<=V;i++)
        {
            int v;
            Scan(v);
            add_edge(V+v,i);
        }
        solve();
    }
    return 0;
}

kosaraju算法——有向图缩点利器

#include"cstdio"
#include"cstring"
#include"algorithm"
#include"vector"
using namespace std;
const int MAXN=4010;
vector<int> G[MAXN];
vector<int> rG[MAXN];
vector<int> vs;
int V,E;
void add_edge(int u,int v)
{
    G[u].push_back(v);
    rG[v].push_back(u);
}
int vis[MAXN];
int cpnt[MAXN];
void dfs(int u)
{
    vis[u]=1;
    for(int i=0;i<G[u].size();i++)
        if(!vis[G[u][i]])    dfs(G[u][i]);
    vs.push_back(u);
}
void rdfs(int u,int k)
{
    vis[u]=1;
    cpnt[u]=k;
    for(int i=0;i<rG[u].size();i++)
        if(!vis[rG[u][i]])    rdfs(rG[u][i],k);
}
void scc()
{
    vs.clear();
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=V+V;i++)
        if(!vis[i])    dfs(i);
    memset(vis,0,sizeof(vis));
    int k=1;
    for(int i=vs.size()-1;i>=0;i--)
        if(!vis[vs[i]])    rdfs(vs[i],k++);
}
int ans[MAXN];
void solve()
{
    memset(ans,0,sizeof(ans));
    scc();
    for(int i=1;i<=V;i++)
    {
        int counter=0;
        for(int j=0;j<G[i].size();j++)
        {
            int v=G[i][j];
            if(cpnt[i]==cpnt[v])    ans[counter++]=v-V;    
        }
        sort(ans,ans+counter);
        printf("%d",counter);
        for(int i=0;i<counter;i++)    printf(" %d",ans[i]);
        printf("
");
    }
}
int main()
{
    while(scanf("%d",&V)!=EOF)
    {
        for(int i=1;i<=V+V;i++)
        {
            G[i].clear();
            rG[i].clear();
        }
            
        for(int i=1;i<=V;i++)
        {
            
            int k;
            scanf("%d",&k);
            while(k--)
            {
                int v;
                scanf("%d",&v);
                add_edge(i,v+V);
            }
        }
        for(int i=1;i<=V;i++)
        {
            int v;
            scanf("%d",&v);
            add_edge(v+V,i);
        }
        
        solve();
    }
    return 0;
}