poj 1144 Network(割点)

题目链接： http://poj.org/problem?id=1144

思路分析：该问题要求求出无向联通图中的割点数目，使用Tarjan算法即可求出无向联通图中的所有的割点，算法复杂度为O(|V| + |E|)；

 

代码如下：

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

const int MAX_N = 100 + 10;
const int MAX_M = 200 + 20;
char str[MAX_M];
vector<int> G[MAX_N];
int pre[MAX_N], is_cut[MAX_N], low[MAX_N];
int dfs_clock;

inline int Min(int a, int b) { return a < b ? a : b; }
int Dfs(int u, int fa)
{
    int lowu = pre[u] = ++dfs_clock;
    int child = 0;

    for (int i = 0; i < G[u].size(); ++i)
    {
        int v = G[u][i];
        if (!pre[v])
        {
            child++;
            int lowv = Dfs(v, u);
            lowu = Min(lowu, lowv);
            if (lowv >= pre[u])
                is_cut[u] = true;
        } else if (pre[v] < pre[u] && v != fa)
            lowu = Min(lowu, pre[v]);
    }
    if (fa < 0 && child == 1)
        is_cut[u] = 0;
    low[u] = lowu;
    return lowu;
}

int main()
{
    int ver_num;

    while (scanf("%d", &ver_num) != EOF && ver_num)
    {
        getchar( );
        int ver_start, ver_next;

        memset(pre, 0, sizeof(pre));
        memset(low, 0, sizeof(low));
        memset(is_cut, 0, sizeof(is_cut));
        for (int i = 0; i < MAX_N; ++i)
            G[i].clear( );
        while (scanf("%d", &ver_start) && ver_start)
        {
            while (getchar() != '\n')
            {
                scanf("%d", &ver_next);
                G[ver_start].push_back(ver_next);
                G[ver_next].push_back(ver_start);
            }
        }
        int cut_ver_count = 0;
        dfs_clock = 0;
        Dfs(1, -1);
        for (int i = 1; i <= ver_num; ++i)
        {
            if (is_cut[i])
                cut_ver_count++;
        }
        printf("%d\n", cut_ver_count);
    }
    return 0;
}