POJ1236 （强连通分量缩点求入度为0和出度为0的分量个数）

Network of Schools

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13804		Accepted: 5507

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

 

题意：

N(2<N<100)各学校之间有单向的网络，每个学校得到一套软件后，可以通过单向网络向周边的学校传输，问题1：初始至少需要向多少个学校发放软件，使得网络内所有的学校最终都能得到软件。2，至少需要添加几条传输线路(边)，使任意向一个学校发放软件后，经过若干次传送，网络内所有的学校最终都能得到软件。

也就是：给定一个有向图，求：

1) 至少要选几个顶点，才能做到从这些顶点出发，可以到达全部顶点

2) 至少要加多少条边，才能使得从任何一个顶点出发，都能到达全部顶点

思路：先求出所有连通分量，将每个连通分量缩成一点，则形成一个有向无环图DAG。为题1的答案就是DAG中入度为0的点个数。问题2等价于在DAG中最少加几条边才能变成强连通。

要为每个入度为0的点加入边，为每个出度为0的点加出边，假设有n个入度为0的点，m个出度为0的点，则答案一定是min(n, m)。另外需要注意的是如果整个图只有一个强连通分支的时候，即缩点后只有一个点，则不需要加边，输出0。

/*
ID: LinKArftc
PROG: 1236.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 110;
const int maxm = 10010;

struct Edge {
    int v, next;
} edge[maxm];

int tot, head[maxn];

void init() {
    tot = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v) {
    edge[tot].v = v;
    edge[tot].next = head[u];
    head[u] = tot ++;
}

int n, m;

int dfn[maxn], low[maxn], ins[maxn], belong[maxn];
int scc, Time;
stack <int> st;
vector <int> vec[maxn];

void tarjan(int u) {
    dfn[u] = low[u] = ++ Time;
    int v;
    st.push(u);
    ins[u] = true;
    for (int i = head[u]; i + 1; i = edge[i].next) {
        v = edge[i].v;
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if (ins[v]) low[u] = min(low[u], low[v]);
    }
    if (low[u] == dfn[u]) {
        scc ++;
        do {
            v = st.top();
            st.pop();
            ins[v] = false;
            vec[scc].push_back(v);
            belong[v] = scc;
        } while (u != v);
    }
}

int indeg[maxn], outdeg[maxn];

int main() {
    //input;
    int v;
    while (~scanf("%d", &n)) {
        init();
        for (int i = 1; i <= n; i ++) {
            while (~scanf("%d", &v) && v) {
                addedge(i, v);
            }
        }
        while (!st.empty()) st.pop();
        for (int i = 1; i <= n; i ++) vec[i].clear();
        memset(dfn, 0, sizeof(dfn));
        memset(ins, 0, sizeof(ins));
        Time = 0;
        scc = 0;
        for (int i = 1; i <= n; i ++) {
            if (!dfn[i]) tarjan(i);
        }
        memset(indeg, 0, sizeof(indeg));
        memset(outdeg, 0, sizeof(outdeg));
        for (int u = 1; u <= n; u ++) {
            for (int i = head[u]; i + 1; i = edge[i].next) {
                v = edge[i].v;
                if (belong[u] == belong[v]) continue;
                outdeg[belong[u]] ++;
                indeg[belong[v]] ++;
            }
        }
        int incnt = 0, outcnt = 0;
        for (int i = 1; i <= scc; i ++) {
            if (indeg[i] == 0) incnt ++;
            if (outdeg[i] == 0) outcnt ++;
        }
        printf("%d
", incnt);
        if (scc == 1) printf("0
");
        else printf("%d
", max(incnt, outcnt));
        
    }

    return 0;
}