Network of Schools（poj1236 Tarjan）

Network of Schools

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27846		Accepted: 10960

Description

A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a list of schools to which it distributes software (the “receiving schools”). Note that if B is in the distribution list of school A, then A does not necessarily appear in the list of school B
You are to write a program that computes the minimal number of schools that must receive a copy of the new software in order for the software to reach all schools in the network according to the agreement (Subtask A). As a further task, we want to ensure that by sending the copy of new software to an arbitrary school, this software will reach all schools in the network. To achieve this goal we may have to extend the lists of receivers by new members. Compute the minimal number of extensions that have to be made so that whatever school we send the new software to, it will reach all other schools (Subtask B). One extension means introducing one new member into the list of receivers of one school.

Input

The first line contains an integer N: the number of schools in the network (2 <= N <= 100). The schools are identified by the first N positive integers. Each of the next N lines describes a list of receivers. The line i+1 contains the identifiers of the receivers of school i. Each list ends with a 0. An empty list contains a 0 alone in the line.

Output

Your program should write two lines to the standard output. The first line should contain one positive integer: the solution of subtask A. The second line should contain the solution of subtask B.

Sample Input

5
2 4 3 0
4 5 0
0
0
1 0

Sample Output

1
2

Source

IOI 1996

解题思路： 题目意思求选则最少的点可以到达所有点      任意一个点可以到达其他所有的点

Tarjan缩点   求入读为0的点  和    第二个把所有的出度为0的点连到入读为0的点即可  那么就是求max(入读为0点的个数和出度为0点的个数)

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn=110;
int n;
int cnt,head[maxn];
int dfn[maxn],low[maxn],scc[maxn],stk[maxn],index=0,sccnum=0,top=0;
struct Edge{
    int v,nxt;
}G[maxn*maxn];
void init(){ cnt=0; memset(head,-1,sizeof(head)); }
void add(int u,int v){
    G[cnt].v=v;
    G[cnt].nxt=head[u];
    head[u]=cnt++;
}

void tarjan(int root){   //tarjan 模板
    if(dfn[root]) return;
    dfn[root]=low[root]=++index;
    stk[++top]=root;

    for(int i=head[root];~i;i=G[i].nxt){
        int v=G[i].v;
        if(!dfn[v]){
            tarjan(v);
            low[root]=min(low[root],low[v]);
        }
        else if(!scc[v]){   //如果还在站内
            low[root]=min(low[root],dfn[v]);
        }
    }
    if(low[root]==dfn[root]){
        sccnum++;
        for(;;){
            int x=stk[top--];
            scc[x]=sccnum;
            if(x==root) break;
        }
    }
}

int in[maxn],out[maxn];
void solve(){
    index=0,sccnum=0,top=0;
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(scc,0,sizeof(scc));
    memset(stk,0,sizeof(stk));
    for(int i=1;i<=n;i++){
        if(!dfn[i]){
            tarjan(i);
        }
    }
    if(sccnum==1){   //特判
        printf("1\n0\n");
        return;
    }
    for(int i=1;i<=sccnum;i++){
        in[i]=0,out[i]=0;
    }
    for(int i=1;i<=n;i++){   //重新建图
        for(int j=head[i];~j;j=G[j].nxt){
            int v=G[j].v;
            if(scc[i]!=scc[v]){
                in[scc[v]]++;
                out[scc[i]]++;
            }
        }
    }
    int res1=0,res2=0;
    for(int i=1;i<=sccnum;i++){
        if(in[i]==0) res1++;
        if(out[i]==0) res2++;
    }
    printf("%d\n%d\n",res1,max(res1,res2));
}

int main(){
    while(scanf("%d",&n)==1){
        init();
        for(int i=1,d1;i<=n;i++){
            while(scanf("%d",&d1)==1&&d1){
                add(i,d1);
            }
        }
        solve();
    }
    return 0;

}