POJ 1236 - Network of Schools(强连通分量)

最新推荐文章于 2024-02-06 19:17:23 发布

晚风中的自由

最新推荐文章于 2024-02-06 19:17:23 发布

阅读量521

点赞数 1

CC 4.0 BY-SA版权

文章标签： poj 强连通

本文链接：https://blog.youkuaiyun.com/u014028317/article/details/45605371

本文介绍了一种基于Tarjan算法的解决方案，用于确定最少数量的学校以确保软件能覆盖整个网络，同时计算需要添加的关系数以实现从任一学校出发都能达到所有学校的网络连接。

Network of Schools

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

Appoint description:

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

Sample Output

1
2

题意：有n个学校，学校之间可以传递信息，为单向传递。

问题一：至少要向几个学校传递原始信息，才能保证所有学校都能收到信息。

问题二：至少要添加多少组关系（每组关系类型如右：a 可以向 b 传递信息），才能保证给任意一个学校原始信息后，其他所有学校都能收到信息。

思路：这道题其实就是一个有n个顶点的有向图，先用 Tarjan 算法缩点，然后分别统计出入度为0 和出度为0 的强连通分量个数 num1 和 num2，那么，问题一的答案就是 num1 , 问题二的答案就是 max（num1 ， num2），但是有特例：当只有一个强连通分量时，问题二的答案就是 0 。

<span style="font-size:18px;">#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;
const int MAXN = 110;
int n;
struct Edge
{
    int v;
    int w;
    int next;
} edge[MAXN*MAXN];
struct shrink_point
{
    int in;
    int out;
    int num;
} sp[MAXN];
int head[MAXN];
int instack[MAXN];
int scc[MAXN];
int LOW[MAXN];
int DFN[MAXN];
stack<int>Q;
int Index, edge_cnt, scc_cnt;

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

void Tarjan(int u)
{
    LOW[u] = DFN[u] = ++Index;
    instack[u] = 1;
    Q.push(u);
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].v;
        if(!DFN[v])
        {
            Tarjan(v);
            if(LOW[u] > LOW[v])
                LOW[u] = LOW[v];
        }
        else if(instack[v] && LOW[u]>DFN[v])
            LOW[u] = DFN[v];
    }
    if(LOW[u] == DFN[u])
    {
        int j;
        scc_cnt++;
        do
        {
            j = Q.top();
            Q.pop();
            instack[j] = 0;
            scc[j] = scc_cnt;
            sp[scc_cnt].num++;
        }
        while(j != u);
    }
}

void solve()
{
    memset(LOW, 0, sizeof(LOW));
    memset(DFN, 0, sizeof(DFN));
    memset(instack, 0, sizeof(instack));
    memset(sp, 0, sizeof(sp));
    while(!Q.empty())
        Q.pop();
    Index = scc_cnt = 0;
    for(int i = 1; i <= n; i++)
        if(!DFN[i])
            Tarjan(i);
    for(int i = 1; i <= n; i++)
        for(int k = head[i]; k != -1; k = edge[k].next)
        {
            int j = edge[k].v;
            if(scc[i] != scc[j])
            {
                sp[scc[i]].out++;
                sp[scc[j]].in++;
            }
        }
    if(scc_cnt == 1)
    {
        printf("%d\n%d\n", 1, 0);
        return;
    }
    int num1 = 0;
    int num2 = 0;
    for(int i = 1; i <= scc_cnt; i++)
    {
        if(sp[i].out == 0)
            num1++;
        if(sp[i].in == 0)
            num2++;
    }
    //printf("%d %d\n", num1, num2);
    printf("%d\n%d\n", num2, max(num1, num2));
}

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    while(cin>>n)
    {
        int v;
        edge_cnt = 0;
        memset(head, -1, sizeof(head));
        memset(edge, 0, sizeof(edge));
        for(int i = 1; i <= n; i++)
        {
            while(cin>>v&&v)
            {
                addedge(i, v);
            }
        }
//        for(int i = 1; i <= n; i++)
//        {
//            printf("%d", i);
//            for(int k = head[i]; k != -1; k = edge[k].next)
//            {
//                printf(" %d", edge[k].v);
//            }
//            printf("\n");
//        }
        solve();
    }
    return 0;
}
</span>