Network of Schools（强连通分量 + Tarjan）

Network of Schools

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10516		Accepted: 4196

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：

       给出 N，代表有 N（2 ~ 100） 个结点，后给出 N 行，每行 i 都给出一系列的节点编号 j ，代表 i 到 j 有一条单向边，直到输入 0 时候退出。输出两个数，第一个数是问最少需要向多少个点发送信息，能使这个信息能到达所有的点，第二个数是需要增加多少条单向边，使这个图任意两点间都能互相到达。

 

       思路：

       强连通分量。先对图进行强连通缩点后变成一个 DAG ，第一个数即找多少个入度为 0 的点。第二个数找的是 入度为 0 的点的个数 和 出度为 0 的点的个数 的最大值。需要注意的是，当这个图本身就是个强连通分量的话要另外讨论，输出的是 1 和 0 。

 

       AC：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>

using namespace std;
const int NMAX = 105;
const int EMAX = 105 * 105;

int n, ind;
int fir[NMAX], Next[EMAX], v[EMAX];

int dfs_clock, scc_cnt;
int cmp[NMAX], low[NMAX], pre[NMAX];
stack<int> s;

int nscc[NMAX][NMAX], in[NMAX], out[NMAX];

void add_edge(int f, int t) {
    v[ind] = t;
    Next[ind] = fir[f];
    fir[f] = ind;
    ++ind;
}

void dfs(int u) {
    pre[u] = low[u] = ++dfs_clock;
    s.push(u);

    for (int e = fir[u]; e != -1; e = Next[e]) {
            if (!pre[ v[e] ]) {
                    dfs( v[e] );
                    low[u] = min(low[u], low[ v[e] ]);
            } else if (!cmp[ v[e] ]) {
                    low[u] = min(low[u], pre[ v[e] ]);
            }
    }

    if (low[u] == pre[u]) {
            ++scc_cnt;
            for (;;) {
                    int x = s.top(); s.pop();
                    cmp[x] = scc_cnt;
                    if (x == u) break;
            }
    }
}

void scc() {
    dfs_clock = scc_cnt = 0;
    memset(cmp, 0, sizeof(cmp));
    memset(pre, 0, sizeof(pre));

    for (int u = 1; u <= n; ++u) {
            if (!pre[u]) dfs(u);
    }
}

void make_scc() {
    memset(nscc, 0, sizeof(nscc));
    memset(in, 0, sizeof(in));
    memset(out, 0, sizeof(out));

    for (int f = 1; f <= n; ++f) {
            for (int e = fir[f]; e != -1; e = Next[e]) {
                    int t = v[e];
                    if (cmp[f] != cmp[t] &&
                        !nscc[ cmp[f] ][ cmp[t] ]) {
                        nscc[ cmp[f] ][ cmp[t] ] = 1;
                        ++in[ cmp[t] ];
                        ++out[ cmp[f] ];
                    }
            }
    }
}

int main()
{
    ind = 0;
    memset(fir, -1, sizeof(fir));

    scanf("%d", &n);
    for (int f = 1; f <= n; ++f) {
            int t;
            while (~scanf("%d", &t) && t) {
                    add_edge(f, t);
            }
    }

    scc();

    make_scc();

    int temp = 0;
    for (int i = 2; i <= n; ++i) {
            if (cmp[i] != cmp[i - 1]) {
                    temp = 1;
                    break;
            }
    }

    if (temp) {
            int in_num = 0, out_num = 0;
            for (int i = 1; i <= scc_cnt; ++i) {
                    if (!out[i]) ++out_num;
                    if (!in[i]) ++in_num;
            }
            printf("%d\n%d\n", in_num, max(out_num, in_num) );
    } else printf("1\n0\n");

    return 0;
}