poj1523

题意：求割点，并求出去掉每个割点图会被分成几个连通分支。

分析：求割点除了tarjan算法，还有一种O(n^2)的算法，就是分别把每个点作为根，进行dfs，看根有几个子结点，如果大于一个则为割点否则不是割点。本题就是观察每个点为根时有几个子结点，去掉该点后的连通分支数等于其子结点数。

View Code

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

#define maxn 1005

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

int id[maxn];
int name[maxn];
int head[maxn];
int n;
int ecount;
bool vis[maxn];
int cnt;

void addedge(int a, int b)
{
    edge[ecount].v = b;
    edge[ecount].next = head[a];
    head[a] = ecount++;
}

int getid(int a)
{
    if (id[a] == -1)
    {
        n++;
        id[a] = cnt;
        name[cnt] = a;
        cnt++;
    }
    return id[a];
}

void make(int a, int b)
{
    int aa = getid(a);
    int bb = getid(b);
    addedge(aa, bb);
    addedge(bb, aa);
}

void dfs(int u)
{
    vis[u] = true;
    for (int i = head[u]; i != -1; i = edge[i].next)
        if (!vis[edge[i].v])
            dfs(edge[i].v);
}

int cal(int a)
{
    int ret = 0;
    memset(vis, 0, sizeof(vis));
    vis[a] = true;
    for (int i = head[a]; i != -1; i = edge[i].next)
        if (!vis[edge[i].v])
        {
            dfs(edge[i].v);
            ret++;
        }
    return ret;
}

int main()
{
    //freopen("t.txt", "r", stdin);
    int a, b, t = 0;
    while (scanf("%d", &a), a)
    {
        t++;
        printf("Network #%d\n", t);
        memset(id, -1, sizeof(id));
        memset(head, -1, sizeof(head));
        n = 0;
        ecount = 0;
        cnt = 0;
        do
        {
            scanf("%d", &b);
            make(a, b);
        } while (scanf("%d", &a), a);
        bool did = false;
        for (int i = 0; i < n; i++)
        {
            int temp = cal(i);
            if (temp > 1)
            {
                did = true;
                printf("  SPF node %d leaves %d subnets\n", name[i], temp);
            }
        }
        if (!did)
            printf("  No SPF nodes\n");
        putchar('\n');
    }
    return 0;
}