POJ1523tarjan求割点

Posted 2021-08-20 hesorchen

题目

POJ 1523

求割点数量以及删除每个割点后连通分量的数量。

代码

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

const int N = 1006;
vector<int> edge[N];

int dfn[N], low[N], ct, cut[N];
void tarjan(int u, bool isroot)
{
    dfn[u] = low[u] = ++ct;
    int tot = 0;
    for (int i = 0; i < edge[u].size(); i++)
    {
        int v = edge[u][i];
        if (!dfn[v])
        {
            tarjan(v, 0);
            low[u] = min(low[u], low[v]);
            tot += low[v] >= dfn[u];
        }
        else
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
    cut[u] = tot - isroot + 1; //删除每个割点会产生的连通分量数量
}
int CA = 0;
int main()
{

    int u, v;
    while (cin >> u, u)
    {
        int n = 0;
        for (int i = 0; i < N; i++)
            edge[i].clear();
        memset(cut, 0, sizeof cut);
        memset(dfn, 0, sizeof dfn);
        memset(low, 0, sizeof low);
        cin >> v;
        n = max(n, max(u, v));
        edge[u].push_back(v);
        edge[v].push_back(u);
        while (cin >> u, u)
        {
            cin >> v;
            n = max(n, max(u, v));
            edge[u].push_back(v);
            edge[v].push_back(u);
        }
        tarjan(1, 1);
        printf("Network #%d\\n", ++CA);
        bool flag = 0;
        for (int i = 1; i <= n; i++)
        {
            if (cut[i] <= 1)
                continue;
            flag = 1;
            printf("  SPF node %d leaves %d subnets\\n", i, cut[i]);
        }
        flag ? flag : puts("  No SPF nodes");
        puts("");
    }
    return 0;
}