poj 1236 Network of Schools 【SCC + 缩点】【最少连接几个点可以直接或间接连接所有点 + 增加最少的边使图强连通】

Network of Schools

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13349		Accepted: 5338

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


题意：给你一个N个点的有向图，问1，最少连接几个点可以直接或间接 连接到所有点。2，最少增加几条边使图强连通。

较简单的SCC题目，这里只说下思路吧。

思路：SCC + 缩点后。统计出入度为0的SCC数sumin和出度为0的SCC数sumout。
答案一就是sumin，答案二就是max(sumin, sumout)。注意只有一个SCC时，输出1 0。

有人请吃饭，O(∩_∩)O~ 回来再刷题。

AC代码：

#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define MAXN 100+10
#define MAXM 10000+10
#define INF 0x3f3f3f3f
using namespace std;
struct Edge
{
    int from, to, next;
};
Edge edge[MAXM];
int head[MAXN], edgenum;
int low[MAXN], dfn[MAXN];
int dfs_clock;
int sccno[MAXN], scc_cnt;
stack<int> S;
bool Instack[MAXN];
int N;
void init()
{
    edgenum = 0;
    memset(head, -1, sizeof(head));
}
void addEdge(int u, int v)
{
    Edge E = {u, v, head[u]};
    edge[edgenum] = E;
    head[u] = edgenum++;
}
void getMap()
{
    int y;
    for(int i = 1; i <= N; i++)
    {
        while(scanf("%d", &y), y)
            addEdge(i, y);
    }
}
void tarjan(int u, int fa)
{
    int v;
    low[u] = dfn[u] = ++dfs_clock;
    S.push(u);
    Instack[u] = true;
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        v = edge[i].to;
        if(!dfn[v])
        {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else if(Instack[v])
            low[u] = min(low[u], dfn[v]);
    }
    if(low[u] == dfn[u])
    {
        scc_cnt++;
        for(;;)
        {
            v = S.top(); S.pop();
            Instack[v] = false;
            sccno[v] = scc_cnt;
            if(v == u) break;
        }
    }
}
void find_cut(int l, int r)
{
    memset(low, 0, sizeof(low));
    memset(dfn, 0, sizeof(dfn));
    memset(sccno, 0, sizeof(sccno));
    memset(Instack, false, sizeof(Instack));
    dfs_clock = scc_cnt = 0;
    for(int i = l; i <= r; i++)
        if(!dfn[i]) tarjan(i, -1);
}
int in[MAXN], out[MAXN];//记录SCC出度和入度
void suodian()
{
    for(int i = 1; i <= scc_cnt; i++) in[i] = out[i] = 0;
    for(int i = 0; i < edgenum; i++)
    {
        int u = sccno[edge[i].from];
        int v = sccno[edge[i].to];
        if(u != v)
            in[v]++, out[u]++;
    }
}
void solve()
{
    find_cut(1, N);
    suodian();
    if(scc_cnt == 1)//一个SCC
    {
        printf("1\n0\n");
        return ;
    }
    int sumin = 0, sumout = 0;//记录入度为0的SCC和出度为0的SCC
    for(int i = 1; i <= scc_cnt; i++)
    {
        if(in[i] == 0)
            sumin++;
        if(out[i] == 0)
            sumout++;
    }
    printf("%d\n%d\n", sumin, max(sumin, sumout));
}
int main()
{
    while(scanf("%d", &N) != EOF)
    {
        init();
        getMap();
        solve();
    }
    return 0;
}